When is $\sum_{i=1}^{m}i^n =0 \bmod \dfrac{m(m+1)}{2}$?

Question

I have tried to creat such formula related to divisibility using triangulair numbers and series , I have got the following problem such that I ask when is $S_n(m)=\sum_{i=1}^{m}i^n =0 \bmod \dfrac{m(m+1)}{2}$ ? , My simple attempt without using any complicated solution, I got that $n$ must be odd positive and integer and $m$ must be arbitrary positive integer , such that I treated only the first case when both $n$ and $m$ are odd we may group the terms of $ S_n(m)$ as follows, and as $n$ is also odd we see by expanding the binomial that : $S_n(m)=m^n+\sum_{i=1}^{(m-1)/2}(i^n+(m-i)^{n})$ this means only that $m | S_n(m)$ but How I can follow this idea to pove that: $m(m+1)/2 | S_n(m)$ if what i have claimed is true ?