jmc

Setting $m=n=0,$ we get $$2f(0)=f(0),$$so $f(0)=0.$

Setting $n=0,$ we get $$2f(m)=\frac{f(2m)}{2}.$$Thus, we can write the given functional equation as $$f(m+n)+f(m-n)=2f(m)+2f(n).$$In particular, setting $n=1,$ we get $$f(m+1)+f(m-1)=2+2f(m),$$so $$f(m+1)=2f(m)-f(m-1)+2$$for all $m\ge 1.$

Then $$\begin{array}{rl}f(2)& =2f(1)-f(0)+2=4,\\ f(3)& =2f(2)-f(1)+2=9,\\ f(4)& =2f(3)-f(2)+2=16,\end{array}$$and so on.

By a straight-forward induction argument, $$f(m)=m^2$$for all nonnegative integers $m.$ Note that this function satisfies the given functional equation, so the sum of all possible values of $f(10)$ is $\boxed{100}.$