Let's determine $x$ and $y$ such that $1234 x y$ is divisible by eight and nine.
The expression is divisible by 9 if

$$
\text { (1) } x+y=0 \quad \text { and } \quad \text { (2) } \quad x+y=17
$$

the expression is divisible by 8 if

$$
10 x+y=8 a \text {. }
$$

Since $10 x+y$ is a two-digit number, it follows that $a \leqq 12$. Subtracting (1) from (3), we get:

$$
9 x=8(a-1)
$$

from which

$$
x=8 \frac{a-1}{9}
$$

$x$ is an integer if $a=10$, or $a=1$. In the first case, $x=8$ and $y=0$, in the second case, $x=0$ and $y=8$. From the equation $x+y=17$, it follows that $x=9, y=8$, or $x=8, y=9$. But neither 98 nor 89 is divisible by 8, so (2) does not provide a solution.

The sought numbers are therefore:

123480 and 123408.

(Pivnyik István, Nyiregyháza.)

(The problem was also solved by: Bartók I., Deutsch E., Deutsch I., Enyedi B., Haar A., Harsányi Z., Jánosy Gy., Kelemen M., Kertész G., König D., Kürti I., Léderer S., Ligeti P., Messer P., Moskovits Zs., Neidenbach E., Popoviciu M., Raab R., Riesz K., Riesz M., Scheiber S., Schlesinger A., Schwarz Gy., Szávay Z., Weisz P.