932. Find two natural numbers if their product is between 120 and 130, and their ratio is between 2 and 3. List all solutions.
Let's denote the required numbers by $k$ and $n$. Then

$$
\left\{\begin{array}{l}
120 < k n < 130 \\
2 < \frac{k}{n} < 3
\end{array}\right.
$$

We multiply these inequalities: $240 < k^2 < 390$. Additionally, we divide the inequalities, in the right part we need to divide 130 by 2 (but not by 3), and in the left part we need to divide 120 by 3 (not by 2). Then $40 < n^2 < 65$.

We obtain the system of inequalities:

$$
\left\{\begin{array}{l}
240 < k^2 < 390 \\
40 < n^2 < 65
\end{array}\right.
$$

This system is a consequence of the original system, as any solution of the first system satisfies the second system, but the second system has solutions that do not satisfy the original system, for example, $k=16, n=7$.

From the second inequality of the last system, it follows that $n=7$ or $n=8$. We consider both cases, finding $k$, and we must do so from the original system of inequalities.

1) Let $n=7$. We get:

$$
\left\{\begin{array}{l}
120 < 7k < 130, \\
2 < \frac{k}{7} < 3
\end{array} \quad \left\{\begin{array}{l}
17 \frac{1}{7} < k < 18 \frac{4}{7} \\
14 < k < 21
\end{array}\right.\right.
$$

From this, $k=18$.

2) Let $n=8$. We will have:

$$
\left\{\begin{array}{l}
120 < 8k < 130, \\
2 < \frac{k}{8} < 3
\end{array} \quad \left\{\begin{array}{l}
15 < k < 16 \frac{1}{4} \\
16 < k < 24
\end{array}\right.\right.
$$

This system has no solutions in integers.

Answer: 18 and 7.