I took it as a challenge to avoid using Solve, which can be slower than a direct assault. If $a$ is the first number in the sum of consecutive positive integers, and $k$ is the count of integers summing to $n$, then $n=k*a+k(k-1)/2$. Solve this for $a=n/k-(k-1)/2$, with bounds $1 \le k \le {\rm Floor}[(\sqrt{8n+1}-1)/2]$. Consider the odd and even divisors of $n$ and $2n$, respectively, to give the following. This includes the $k=1$ case of the sum of one number, just $n$.

CoreyPartition[n_] := Block[{bound = Floor[(Sqrt[1 + 8 n] - 1)/2], oddk, evenk, k, e}, oddk = Pick[#,OddQ[#]] &[Pick[#, UnitStep[bound - #], 1]&[Divisors[n]]]; evenk = Pick[#, UnitStep[bound - #], 1] &[Divisors[2 n]]; e = IntegerExponent[n, 2]; evenk = Pick[evenk, IntegerExponent[evenk, 2], 1 + e]; k = Reverse[Sort[Join[oddk, evenk]]]; Transpose[{n/k - (k - 1)/2, n/k + (k - 1)/2}]] 

A slight modification to f[n] for cases with no solution, such as $n=1,2,4,8,\ldots$.

f[n_] := If[# == {}, {}, {a, b} /. #]&[ Solve[(a + b) (b - a + 1)/2 == n && 0 < a < n && 0 < b < n, {a, b}, Integers]] 

The two solutions agree, for example:

Sum[Total[Most[CoreyPartition[n]] - f[n]], {n, 1, 200}] 

However, testing with large integers such as n=10^40 showed CoreyPartition[n] was over 200 times faster.

I took it as a challenge to avoid using Solve, which can be slower than a direct assault. If $a$ is the first number in the sum of consecutive positive integers, and $k$ is the count of integers summing to $n$, then $n=k*a+k(k-1)/2$. Solve this for $a=n/k-(k-1)/2$, with bounds $1 \le k \le {\rm Floor}[(\sqrt{8n+1}-1)/2]$. Consider the odd and even divisors of $n$ and $2n$, respectively, to give the following. This includes the $k=1$ case of the sum of one number, just $n$.

CoreyPartition[n_] := Block[{bound = Floor[(Sqrt[1 + 8 n] - 1)/2], oddk, evenk, k, e}, oddk = Pick[#,OddQ[#]] &[Pick[#, UnitStep[bound - #], 1]&[Divisors[n]]]; evenk = Pick[#, UnitStep[bound - #], 1] &[Divisors[2 n]]; e = IntegerExponent[n, 2]; evenk = Pick[evenk, IntegerExponent[evenk, 2], 1 + e]; k = Reverse[Sort[Join[oddk, evenk]]]; Transpose[{n/k - (k - 1)/2, n/k + (k - 1)/2}]] 

A slight modification to f[n] for cases with no solution, such as $n=1,2,4,8,\ldots$.

f[n_] := If[# == {}, {}, {a, b} /. #]&[ Solve[(a + b) (b - a + 1)/2 == n && 0 < a < n && 0 < b < n, {a, b}, Integers]] 

The two solutions agree, for example:

Sum[Total[Most[CoreyPartition[n]] - f[n]], {n, 1, 200}] 

However, testing with large integers such as n=10^40 showed CoreyPartition[n] was over 200 times faster.

Update

Given DivisorPairs from Mr.Wizard:

DivisorPairs[n_] := Thread[{#, Reverse[#]}][[ ;; Ceiling[Length[#]/2]]] &[Divisors[n]] 

there is the following one-liner which is twice as fast as CoreyPartition.

CoreyFastPartition[n_] := Reverse[Pick[#, Total[Mod[#, 4], {2}], 2]] &[ DivisorPairs[8 n]] /. {r_Integer, s_Integer} -> {(s - r + 2)/4, (s + r - 2)/4} 

This code returns the trivial solution {n,n}, which may be removed with Most.