3. Determine all positive integers $m, n$ such that $n$ divides $2 m-1$ and $m$ divides $2 n-1$.
3. We are looking for pairs of positive integers $m$ and $n$ for which there exist positive integers $k$ and $l$ such that

$$
2 m-1=k n \quad \text { and } \quad 2 n-1=l m
$$

Let's consider the numbers $k, l$ as parameters and solve the system of linear equations (1) for the unknowns $m, n$. If, for example, we add $k$ times the first equation to $k$ times the second equation, we eliminate the unknown $n$ and after rearrangement we get the first of the equations

$$
(4-k l) m=k+2 \quad \text { and } \quad(4-k l) n=l+2 \text {; }
$$

the second equation is obtained analogously. Since the right-hand sides of equations (2) are positive, it follows from the form of the left-hand sides that $4-k l>0$ or $k l<4$. From the inequalities $2 m-1 \geq k n$ and $2 n-1 \geq l m$, we get $2 m-1 \geq 2 n$ and $2 n-1 \geq 2 m$, respectively, or $2 m > 2 n$. Therefore, at least one of the equalities $m=2 n-1$ or $n=2 m-1$ must hold. If $m=2 n-1$, then $2 m-1=4 n-3$ and the remaining condition $n \mid 2 m-1$ becomes $n \mid 4 n-3$ or $n \mid 3$. This is satisfied only by the numbers $n=1$ and $n=3$, which correspond to the values $m=1$ and $m=5$ according to the formula $m=2 n-1$. The second case, where $n=2 m-1$, differs from the first only in the exchange of the roles of $m$ and $n$, so in this case we get another suitable pair $m=3$ and $n=5$.

Second approach. Considering the symmetry, we can assume that $m \leq n$, from which it follows that $2 m-1 \leq 2 n-1 < 2 n$. The number $2 m-1$ is thus a multiple of the number $n$ less than $2 n$, so it must be the number $n$ itself. Thus, we have derived the equality $2 m-1=n$. The rest of the reasoning is the same as in the first approach.

Third approach. Notice that the number $2 m+2 n-1$ is divisible by both numbers $m$ and $n$, which are also coprime, since, for example, the number $m$ is a divisor of the number $2 n-1$, which is clearly coprime with the number $n$. Therefore, the number $2 m+2 n-1$ is divisible by the product $m n$, so the inequality $m n \leq 2 m+2 n-1$ or $(m-2)(n-2) \leq 3$ holds. From this, it follows that both numbers $m, n$ cannot be greater than 3; considering the symmetry, we only need to consider the case $m \leq 3$. For $m=1$, the condition $n \mid 2 m-1$ implies $n=1$, for $m=2$ the condition $m \mid 2 n-1$ is not satisfied, and for $m=3$ we have the conditions $3 \mid 2 n-1$ and $n \mid 5$, which are satisfied only by $n=5$.

For a complete solution, 6 points are awarded. For various inequality considerations, award 1 to 4 points, for example, 1 point for distinguishing the cases $m \leq n$ and $n \leq m$, 2 points for effectively using the implication $(a \neq b \wedge a \mid b) \Rightarrow b \geq 2 a$, 4 points for proving the insight that at least one of the equalities $m=2 n-1, n=2 m-1$ must hold. If the solver finds all three suitable pairs but cannot prove that no other solutions exist, award 1 point.