Let's find all pairs of natural numbers where the product of the two numbers is 10 times the absolute value of their difference.
Let $a$ and $b$ be suitable numbers, where $a > b$. Thus, the requirement is:

$$
a b = 10(a - b)
$$

and through appropriate manipulation,

$$
a = \frac{-10 b}{b - 10} = -10 - \frac{100}{b - 10} = \frac{100}{10 - b} - 10
$$

From this, we get a positive integer for $a$ if and only if the fraction in the last form is an integer greater than 10, which means $10 - b$ is a positive divisor of 100 that is less than 10, and in every such case, $b$ will also be a natural number. Therefore, the solutions are:

$$
\begin{array}{rrrr}
10 - b = 1, & 2, & 4, & 5 \\
b = 9, & 8, & 6, & 5 \\
a = 90, & 40, & 15, & 10
\end{array}
$$

By swapping these pairs, we get 4 new solutions, as the absolute value of the difference remains unchanged after the swap. Zoltán Szabó (Budapest, I. István Gymn., II. o. t.)

Note. We can also achieve the goal by collecting the terms of (1) on one side and adding 100 to both sides, so that the multi-term side can be factored:

$$
10 a - 10 b - a b + 100 = (10 - b)(a + 10) = 100
$$