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.
II. Solution: Let $S=\frac{m^{3}+n^{3}-m^{2} n^{2}}{(m+n)^{2}}$, then $S=\frac{(m+n)\left((m+n)^{2}-3 m n\right)-m^{2} n^{2}}{(m+n)^{2}}=m+n-\frac{3 m n}{m+n}-\left(\frac{m n}{m+n}\right)^{2}$, where $m, n \in N_{+}$, let $\frac{m n}{m+n}=\frac{p}{q}, \quad(p, q)=1, p, q \in N_{+}, \quad$ then $S=m+n-\frac{3 p}{q}-\frac{p^{2}}{q^{2}}=m+n-\frac{p^{2}+3 p q}{q^{2}} \in Z$,
therefore $q \mid p^{2}$, then $q=1$, hence $m+n \mid m n$, then we have $n-\frac{m n}{m+n}=\frac{n^{2}}{m+n} \in N$, thus $n^{2} \geq m+n$, i.e., $n^{2}-m \geq n$. From the problem, $S \geq 0$, we know $n^{3} \geq m^{2} n^{2}-m^{3}=m^{2}\left(n^{2}-m\right) \geq m^{2} n$, so $n \geq m$. By the symmetry of $m, n$, similarly, we have $m \geq n$, thus we get $m=n$, and $\frac{m n}{m+n}=\frac{m}{2} \in Z$, so $2 \mid m$, and by $S=\frac{2 m-m^{2}}{4} \geq 0$, solving gives $0<m \leq 2$, hence $m=2$, thus $n=2$. In summary, $(m, n)=(2,2)$ is the unique positive integer solution.