4. Natural numbers $m, n$ are such that the fraction $\frac{m}{n}$ is irreducible, while the fraction $\frac{4 m+3 n}{5 m+2 n}$ is reducible. By which natural numbers can it be reduced?
# Answer: 7

Let the numbers $4 m+3 n$ and $5 m+2 n$ have a common divisor $d>1$. Then there exist natural numbers $a$ and $b$ such that the system holds:

$$
\left\{\begin{array}{l}
4 m+3 n=a d \\
5 m+2 n=b d
\end{array}\right.
$$

From this system, we can express $m$ and $n$: $7 n=d(5 a-4 b), 7 m=d(3 b-2 a)$. If 7 does not divide $d$, then $m$ and $n$ are divisible by $d>1$, which contradicts the irreducibility of the fraction $\frac{m}{n}$. Therefore, $d=7$.