3. Determine all pairs $(m, n)$ of positive integers for which the number $4(m n+1)$ is divisible by the number $(m+n)^{2}$.
3. First, let us realize that for every pair $(m, n)$ of positive integers that satisfy the problem, the pair $(n, m)$ also satisfies it. Therefore, without loss of generality, we can assume that $m \geqq n$.

If the positive integer $A=(m+n)^{2}$ divides the positive integer $B=4(m n+1)$, it is necessary that

$$
(m+n)^{2} \leqq 4(m n+1) \quad \text { or } \quad(m-n)^{2} \leqq 4
$$

Thus, $0 \leqq m-n \leqq 2$. Therefore, one of the following three cases will occur:

$\triangleright m=n$, then $\bar{A}=4 n^{2}, B=4 n^{2}+4$ and $A$ divides $B$ if and only if $4 n^{2}$ divides 4, i.e., $n=1$.

We get one solution $(m, n)=(1,1)$.

$\triangleright m=n+1$, then $A=4 n^{2}+4 n+1, B=4 n^{2}+4 n+4=A+3$. The number $A$ divides $B$ if and only if $4 n^{2}+4 n+1$ divides 3. However, for positive integers $n$, $4 n^{2}+4 n+1 \geqq 4+4+1=9$, so in this case, the problem has no solution.

$\triangleright m=n+2$, then $A=4 n^{2}+8 n+4, B=4 n^{2}+8 n+4$. We see that $A=B$, thus every pair $(m, n)=(n+2, n)$ of positive integers is a solution to the given problem.

Conclusion. The pair $(1,1)$ satisfies the problem, and further (considering the symmetry of the unknowns $m, n$), every pair $(n+2, n)$ and $(m, m+2)$, where $m$ and $n$ are any positive integers, also satisfies the problem. For a complete solution, award 6 points, of which 3 points are for deriving the inequality $(m-n)^{2} \leqq 4$. Additionally, award 1 point for finding the solution $(1,1)$ (deduct 1 point for its omission in an otherwise complete solution) and 2 points for finding all other solutions. For merely listing all solutions without any justification, award no more than 2 points.