Among the poor, 120 K was distributed. If the number of the poor was 10 less, then each would receive exactly as much more as they would receive less if the number of the poor was 20 more. How many poor were there?
Let the number of the poor be denoted by $x$. According to the problem,

$$
\frac{120}{x-10}-\frac{120}{x}=\frac{120}{x}-\frac{120}{x+20}
$$

or

$$
\frac{x-(x-10)}{(x-10) x}=\frac{x+20-x}{x(x+20)}
$$

which simplifies to

$$
\frac{1}{x-10}=\frac{2}{x+20}
$$

from which

$$
x=40
$$

(József Schlesinger, Györ.)

Number of solutions: 31.