Iranian Mathematical Olympiad

It is known that if $n$ divides $a-b$, then $a^n-b^n$ is divisible by $n^2$. Now, choose $z\equiv x(\mathrm{mod}n)$ and $y\equiv t(\mathrm{mod}n)$. Then $$x^n+y^n-(z^n+t^n)=(x^n-z^n)+(y^n-t^n)\equiv 0(\mathrm{mod}{n}^2)$$ Thus, for $z,t$ belonging to $\{0,1,\dots ,n-1\}$, we must calculate maximal number of different values of $z^n+t^n(\mathrm{mod}{n}^2)$. Then, there are $x,y$ such that $x^n+y^n\equiv z^n+t^n(\mathrm{mod}{n}^2)$. These numbers are at most $\frac{n(n+1)}{2}$. Therefore, there are at least $n^2-\frac{n(n+1)}{2}=\frac{n(n-1)}{2}$ different values for $m$.