Let's find three-digit numbers that are 34 times larger than the sum of their digits.
If $x, y$ and $z$ are the digits of the sought number, then

$$
100 x+10 y+z=34(x+y+z)
$$

or

$$
22 x-8 y-11 z=0
$$

from which

$$
2 x-z=\frac{8}{11} y
$$

Since the left side is an integer, the right side must also be an integer, but $y<10$, so it is necessary that $y=0$.

Thus

$$
2 x=z \text {. }
$$

The condition is therefore met by those numbers whose middle digit is $0$, and the last digit is twice the first. Such numbers are:

$102,204,306,408$.

(Kürti Imre, Eger.)

The problem was also solved by: Ádámffy E., Bartók I., Demjén E., Deutsch E., Deutsch I., Enyedi B., Haar A., Harsányi Z., Hirschfeld Gy., Jánosy Gy., Kertész G., Kiss J., König D., Liebner A., Messer P., Moskovits Zs., Pám M., Pivnyik I., Popoviciu M., Preisich G., Prékopa D., Rássy P., Riesz K., Scheiber S., Schlesinger O., Schuster Gy., Schwarz Gy., Sonnenfeld S., Szántó H., Szücs A., Tóth B., Weisz P.