# Python 2, 94 bytes

<!-- language-all: lang-python -->

    n=input()
    r=int((2*n)**.5)
    while r:
     if~r%2*r/2==n%r:print range(n/r-~-r/2,n/r-~r/2);r=1
     r-=1

Input is taken from stdin. This solution is suitable for very large inputs.

This iterates over the possible solution lengths, _r_, having _r ≤ √(2n)_, and checks for a solution explicitly. In order for a solution to exist, if _r_ is odd, _n mod r_ must be zero, and if _r_ is even, _n mod r_ must be _r/2_.

---

**Sample Usage**

    $ echo 8192 | python sum-con-int.py
    [8192]

    $ echo 1000002 | python sum-con-int.py
    [83328, 83329, 83330, 83331, 83332, 83333, 83334, 83335, 83336, 83337, 83338, 83339]

    $ echo 1000000006 | python sum-con-int.py
    [250000000, 250000001, 250000002, 250000003]

I've deliberately choosen examples with relatively small outputs.