JavaScript (ES6), 259 257 248 241 bytes
Returns a Promise object containing \n separated groups, in ascending order of reputation, of profile links for users with the same rep, with each link separated by a comma. As the question only states that users with 200 reputation or less can be ignored (rather than should be ignored), this solution includes a couple of users who do have <200 rep.
f=(x=95,s="",c=0)=>fetch(`//api.stackexchange.com/users?page=${x}&site=codegolf`).then(a=>a.json()).then(j=>(i=j.items.reverse()).map(x=>(r=x.reputation)>c?(c=r,t.length-1&&(s+=t+`
`),t=[x.link]):t.push(x.link),t=[i[0].link])&--p?f(p,s,c):s)
Caveat
Between this challenge, "Martin vs. Dennis" and a couple of others, I've managed to get myself (temporarily) booted from the API for a few hours! I was, therefore, unable to run my final tests on this but it should be working (although, I suspect it might not include users from page 1 in the output). There's probably a bit more I can golf off it, so I'll come back to it when my ban is lifted.
Try it
WARNING: The first Snippet below will submit 95(!) requests to the SE API - running it a couple of times may result in you also having your access to the API temporarily revoked. If you'd prefer to test this with a reduced number of API calls and, therefore, a reduced result set, change the value of the first parameter to something much lower than 95. Alternatively, you can test it with no API calls, using the second batch of sample data from the question by using the second Snippet below, which returns usernames instead of profile links (for now).
f=(x=95,s="",c=0)=>fetch(`//api.stackexchange.com/users?page=${x}&site=codegolf`).then(a=>a.json()).then(j=>(i=j.items.reverse()).map(x=>(r=x.reputation)>c?(c=r,t.length-1&&(s+=t+`
`),t=[x.link]):t.push(x.link),t=[i[0].link])&--p?f(p,s,c):s)
f().then(console.log)
f=(x=95,s="",c=0)=>Promise.resolve(JSON.parse(`{"items":[{"display_name":"John Skeet","reputation":1000000},{"display_name":"Martin Ender","reputation":10000},{"display_name":"Dennis","reputation":10000},{"display_name":"xnor","reputation":5000},{"display_name":"sp3000","reputation":3000},{"display_name":"Digital Trauma","reputation":2000},{"display_name":"Luis Mendo","reputation":2000},{"display_name":"Helka Homba","reputation":2000}]}`)).then(j=>(i=j.items.reverse()).map(y=>(r=y.reputation)>c?(c=r,t.length-1&&(s+=t+`
`),t=[y.display_name]):t.push(y.display_name),t=[i[0].display_name])&--x?f(x,s,c):s)
f().then(console.log)
