946. Find two natural numbers whose product does not exceed their sum. List all solutions.
Let's denote the required numbers by $x$ and $y$. Then

$$
x y \leqslant x+y .
$$

Let $x \leqslant y$. We obtain:

$$
x y \leqslant x+y \leqslant y+y=2 y
$$

From this,

$$
x y \leqslant 2 y, \quad x \leqslant 2
$$

Now let's consider two cases.

1) Suppose $x=1$.

The original inequality becomes:

$$
y \leqslant 1+y
$$

This inequality holds for any natural $y$.

2) Suppose $x=2$.

Then

$$
2 y \leqslant 2+y, \quad y \leqslant 2
$$

Since in this case the inequality $y \geqslant x=2$ must hold, we have $y=2$.

Answer: one number is 1, the other is any natural number; $2,2$.

Note that in solving problem 946, we applied a method similar to the one used in solving some equations in natural numbers (see problems $894-899$ from $\S 20$ ).