Berkeley Math Circle

Solution:
Adding $f(m)$ to both sides, we get $f(m+f(n))=n+f(m)$. Swapping $m$ and $n$ gives $m+f(n)=f(n+f(m))=f(f(m+f(n)))$. By fixing $n$ and varying $m$, we can get $m+f(n)$ to be any integer $x$. Thus, $x=f(f(x))$.
Plugging in $1$ for $m$ and $f(x)$ for $n$ to the original equation then gives $f(x)+f(1)=f(1+f(f(x)))=f(1+x)$, so $f(x)$ is linear, of the form $c x$ for some constant $c$.

Plugging in $0$ for $m$ then gives $n=f(f(n))-f(0)=c^{2} n$, so $c^{2}= \pm 1$. It is easily seen that both $f(x)=x$ and $f(x)=-x$ indeed work.