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#Python 2.7, 463 bytes ###thanks to NoOneIsHere for pointing out additional improvements!

Try it online!

from itertools import permutations
l,m=4,0
p=list(permutations([(1,1),(2,2),(3,3)]))
def d(s,e):
    d=0
    while s!=e:
        x = 1 if s[0]<e[0] else -1 if s[0]>e[0] else 0
        y=1 if s[1]<e[1] else -1 if s[1]>e[1] else 0
        s=(s[0]+x,s[1]+y)
        d += (1, 1.5)[x and y]
    return d
for i in range(len(p)):
    a,c=l-d((0,0),(p[i][0])),1
    for j in range(len(p[i])-1):
        a-=d(p[i][j],p[i][j+1])
        c+=(0,1)[a>0]
    m=max(c,m)
print m

The function calculates the distance between two points according to the given rules, the loop iterates through all permutations of the input and calculates the distance, if the distance is lesser than the candle lifespan it subtracts it and adds the place to the counter, if that counter is greater than the current max it substitutes it.

from itertools import permutations
lifespan, max_amount = 4, 0
perms = list(permutations([(1,1), (2,2), (3,3)]))
    
def distance(start_pos, end_pos):
    distance = 0
    while start_pos != end_pos:
        mod_x = 1 if start_pos[0] < end_pos[0] else -1 if start_pos[0] > end_pos[0] else 0
        mod_y = 1 if start_pos[1] < end_pos[1] else -1 if start_pos[1] > end_pos[1] else 0
        start_pos = (start_pos[0] + mod_x, start_pos[1] + mod_y)
        distance += (1, 1.5)[mod_x and mod_y]
    return distance
    
for i in range(len(perms)):
    lifespan_local, current = lifespan - distance((0,0), (perms[i][0])), 1
    for j in range(len(perms[i]) - 1):
        lifespan_local -= distance(perms[i][j], perms[i][j + 1])
        current += (0, 1)[lifespan_local > 0]
    max_amount = max(current, max_amount)
print max_amount

#Python 2.7, 422 bytes ###thanks to NoOneIsHere for pointing out additional improvements! ###thanks to edc65 for noting not to save the list but use iterators instead!

Try it online!

from itertools import permutations
def d(s,e):
    d=0
    while s!=e:
        x=1 if s[0]<e[0] else -1 if s[0]>e[0] else 0
        y=1 if s[1]<e[1] else -1 if s[1]>e[1] else 0
        s=(s[0]+x,s[1]+y)
        d+=(1,1.5)[x and y]
return d
l,m=4,0
for o in permutations([(1,1),(2,2),(3,3)]):
    a,c=l-d((0,0),o[0]),1
    for j in range(len(o)-1):
        a-=d(o[j],o[j+1])
        c+=(0,1)[a>0]
    m=max(c,m)
print m

The function calculates the distance between two points according to the given rules, the loop iterates through all permutations generated by the generator of the input and calculates the distance, if the distance is lesser than the candle lifespan it subtracts it and adds the place to the counter, if that counter is greater than the current max it substitutes it.

from itertools import permutations

def distance(start_pos, end_pos):
    distance = 0
    while start_pos != end_pos:
        mod_x = 1 if start_pos[0] < end_pos[0] else -1 if start_pos[0] > end_pos[0] else 0
        mod_y = 1 if start_pos[1] < end_pos[1] else -1 if start_pos[1] > end_pos[1] else 0
        start_pos = (start_pos[0] + mod_x, start_pos[1] + mod_y)
        distance += (1, 1.5)[mod_x and mod_y]
    return distance 

lifespan, max_amount = 4, 0
for item in permutations([(1,1), (2,2), (3,3)]):
    lifespan_local, current = lifespan - distance((0,0), item[0]), 1
    for j in range(len(item) - 1):
        lifespan_local -= distance(item[j], item[j + 1])
        current += (0, 1)[lifespan_local > 0]
    max_amount = max(current, max_amount)
print max_amount

#Python 2.7, 484 bytes

Try it online!

from itertools import permutations
l,m=4,0
p=list(permutations([(1,1),(2,2),(3,3)]))
def d(s,e):
    d=0
    while s!=e:
        x = 1 if s[0]<e[0] else -1 if s[0]>e[0] else 0
        y=1 if s[1]<e[1] else -1 if s[1]>e[1] else 0
        s=(s[0]+x,s[1]+y)
        d += (1, 1.5)[x and y]
    return d
for i in range(len(p)):
    ll,current=l-d((0,0),(p[i][0])),1
    for j in range(len(p[i])-1):
        ll-=d(p[i][j],p[i][j+1])
        current+=(0,1)[ll>0]
    m=max(current,m)
print m

#Explanation:

The function calculates the distance between two points according to the given rules, the loop iterates through all permutations of the input and calculates the distance, if the distance is lesser than the candle lifespan it subtracts it and adds the place to the counter, if that counter is greater than the current max it substitutes it.

#ungolfed:

from itertools import permutations
lifespan, max_amount = 4, 0
perms = list(permutations([(1,1), (2,2), (3,3)]))
    
def distance(start_pos, end_pos):
    distance = 0
    while start_pos != end_pos:
        mod_x = 1 if start_pos[0] < end_pos[0] else -1 if start_pos[0] > end_pos[0] else 0
        mod_y = 1 if start_pos[1] < end_pos[1] else -1 if start_pos[1] > end_pos[1] else 0
        start_pos = (start_pos[0] + mod_x, start_pos[1] + mod_y)
        distance += (1, 1.5)[mod_x and mod_y]
    return distance
    
for i in range(len(perms)):
    lifespan_local, current = lifespan - distance((0,0), (perms[i][0])), 1
    for j in range(len(perms[i]) - 1):
        lifespan_local -= distance(perms[i][j], perms[i][j + 1])
        current += (0, 1)[lifespan_local > 0]
    max_amount = max(current, max_amount)
print max_amount

#Python 2.7, 463 bytes ###thanks to NoOneIsHere for pointing out additional improvements!

Try it online!

from itertools import permutations
l,m=4,0
p=list(permutations([(1,1),(2,2),(3,3)]))
def d(s,e):
    d=0
    while s!=e:
        x = 1 if s[0]<e[0] else -1 if s[0]>e[0] else 0
        y=1 if s[1]<e[1] else -1 if s[1]>e[1] else 0
        s=(s[0]+x,s[1]+y)
        d += (1, 1.5)[x and y]
    return d
for i in range(len(p)):
    a,c=l-d((0,0),(p[i][0])),1
    for j in range(len(p[i])-1):
        a-=d(p[i][j],p[i][j+1])
        c+=(0,1)[a>0]
    m=max(c,m)
print m

#Explanation:

The function calculates the distance between two points according to the given rules, the loop iterates through all permutations of the input and calculates the distance, if the distance is lesser than the candle lifespan it subtracts it and adds the place to the counter, if that counter is greater than the current max it substitutes it.

#ungolfed:

from itertools import permutations
lifespan, max_amount = 4, 0
perms = list(permutations([(1,1), (2,2), (3,3)]))
    
def distance(start_pos, end_pos):
    distance = 0
    while start_pos != end_pos:
        mod_x = 1 if start_pos[0] < end_pos[0] else -1 if start_pos[0] > end_pos[0] else 0
        mod_y = 1 if start_pos[1] < end_pos[1] else -1 if start_pos[1] > end_pos[1] else 0
        start_pos = (start_pos[0] + mod_x, start_pos[1] + mod_y)
        distance += (1, 1.5)[mod_x and mod_y]
    return distance
    
for i in range(len(perms)):
    lifespan_local, current = lifespan - distance((0,0), (perms[i][0])), 1
    for j in range(len(perms[i]) - 1):
        lifespan_local -= distance(perms[i][j], perms[i][j + 1])
        current += (0, 1)[lifespan_local > 0]
    max_amount = max(current, max_amount)
print max_amount

Python 2.7, 484 bytes

from itertools import permutations
l,m=4,0
p=list(permutations([(1,1),(2,2),(3,3)]))
def d(s,e):
    d=0
    while s!=e:
        x = 1 if s[0]<e[0] else -1 if s[0]>e[0] else 0
        y=1 if s[1]<e[1] else -1 if s[1]>e[1] else 0
        s=(s[0]+x,s[1]+y)
        d += (1, 1.5)[x and y]
    return d
for i in range(len(p)):
    ll,current=l-d((0,0),(p[i][0])),1
    for j in range(len(p[i])-1):
        ll-=d(p[i][j],p[i][j+1])
        current+=(0,1)[ll>0]
    m=max(current,m)
print m

Explanation:

The function calculates the distance between two points according to the given rules, the loop iterates through all permutations of the input and calculates the distance, if the distance is lesser than the candle lifespan it subtracts it and adds the place to the counter, if that counter is greater than the current max it substitutes it.

ungolfed:

from itertools import permutations
lifespan, max_amount = 4, 0
perms = list(permutations([(1,1), (2,2), (3,3)]))
    
def distance(start_pos, end_pos):
    distance = 0
    while start_pos != end_pos:
        mod_x = 1 if start_pos[0] < end_pos[0] else -1 if start_pos[0] > end_pos[0] else 0
        mod_y = 1 if start_pos[1] < end_pos[1] else -1 if start_pos[1] > end_pos[1] else 0
        start_pos = (start_pos[0] + mod_x, start_pos[1] + mod_y)
        distance += (1, 1.5)[mod_x and mod_y]
    return distance
    
for i in range(len(perms)):
    lifespan_local, current = lifespan - distance((0,0), (perms[i][0])), 1
    for j in range(len(perms[i]) - 1):
        lifespan_local -= distance(perms[i][j], perms[i][j + 1])
        current += (0, 1)[lifespan_local > 0]
    max_amount = max(current, max_amount)
print max_amount

#Python 2.7, 484 bytes

Try it online!

from itertools import permutations
l,m=4,0
p=list(permutations([(1,1),(2,2),(3,3)]))
def d(s,e):
    d=0
    while s!=e:
        x = 1 if s[0]<e[0] else -1 if s[0]>e[0] else 0
        y=1 if s[1]<e[1] else -1 if s[1]>e[1] else 0
        s=(s[0]+x,s[1]+y)
        d += (1, 1.5)[x and y]
    return d
for i in range(len(p)):
    ll,current=l-d((0,0),(p[i][0])),1
    for j in range(len(p[i])-1):
        ll-=d(p[i][j],p[i][j+1])
        current+=(0,1)[ll>0]
    m=max(current,m)
print m

#Explanation:

The function calculates the distance between two points according to the given rules, the loop iterates through all permutations of the input and calculates the distance, if the distance is lesser than the candle lifespan it subtracts it and adds the place to the counter, if that counter is greater than the current max it substitutes it.

#ungolfed:

from itertools import permutations
lifespan, max_amount = 4, 0
perms = list(permutations([(1,1), (2,2), (3,3)]))
    
def distance(start_pos, end_pos):
    distance = 0
    while start_pos != end_pos:
        mod_x = 1 if start_pos[0] < end_pos[0] else -1 if start_pos[0] > end_pos[0] else 0
        mod_y = 1 if start_pos[1] < end_pos[1] else -1 if start_pos[1] > end_pos[1] else 0
        start_pos = (start_pos[0] + mod_x, start_pos[1] + mod_y)
        distance += (1, 1.5)[mod_x and mod_y]
    return distance
    
for i in range(len(perms)):
    lifespan_local, current = lifespan - distance((0,0), (perms[i][0])), 1
    for j in range(len(perms[i]) - 1):
        lifespan_local -= distance(perms[i][j], perms[i][j + 1])
        current += (0, 1)[lifespan_local > 0]
    max_amount = max(current, max_amount)
print max_amount