II. (40 points) Find all positive integers $m, n$ such that $\frac{m^{3}+n^{3}-m^{2} n^{2}}{(m+n)^{2}}$ is a non-negative integer.
Let $S=\frac{m^{3}+n^{3}-m^{2} n^{2}}{(m+n)^{2}}$.
Then $S=\frac{(m+n)\left((m+n)^{2}-3 m n\right)-m^{2} n^{2}}{(m+n)^{2}}$
$$
=(m+n)-\frac{3 m n}{m+n}-\left(\frac{m n}{m+n}\right)^{2} \text {. }
$$

Since $m, n$ are positive integers, let
$$
\frac{m n}{m+n}=\frac{q}{p}((p, q)=1) \text {. }
$$

Thus, $S=(m+n)-\frac{3 q}{p}-\frac{q^{2}}{p^{2}}$
$$
=(m+n)-\frac{3 p q+q^{2}}{p^{2}} \text {. }
$$

Since $S$ is an integer, it follows that $p \mid q^{2}$.
Then $p=1 \Rightarrow(m+n) \mid m n$.
Also, $\frac{n^{2}}{m+n}=n-\frac{m n}{m+n}$ is an integer, and $\frac{n^{2}}{m+n}>0$,
so $n^{2} \geqslant m+n \Rightarrow n^{2}-m \geqslant n$.
By $S \geqslant 0$, we know $m^{3}+n^{3}-m^{2} n^{2} \geqslant 0$.
Thus, $n^{3} \geqslant m^{2} n^{2}-m^{3}$
$$
\begin{array}{l}
=m^{2}\left(n^{2}-m\right) \geqslant m^{2} n \\
\Rightarrow n^{2} \geqslant m^{2} \Rightarrow n \geqslant m .
\end{array}
$$

By the symmetry of $m, n$, similarly, $m \geqslant n$.
Therefore, $m=n$.
Substituting into $\frac{m n}{m+n}$, we have
$$
\begin{array}{l}
\left.\frac{m}{2} \in \mathbf{Z}_{+} \Rightarrow 2 \right\rvert\, m . \\
\text { Also, } S=2 m-\frac{3}{2} m-\left(\frac{m}{2}\right)^{2} \geqslant 0 \\
\Rightarrow 0<m \leqslant 2,
\end{array}
$$

Thus, $m=2$.
Therefore, the only pair $(m, n)$ that satisfies the conditions is $(2,2)$.