Let's find the numbers between 1000 and 4000 that, when divided by 11, 13, and 19, respectively leave remainders of 2, 12, and 18.
If the sought number is $x$, then

$$
x=11 y+2=13 z+12=19 v+18
$$

so

$$
11 y+2=13 z+12
$$

and

$$
11 y+2=19 v+18
$$

From (1)

$$
y=13 m-5
$$

From (2)

$$
y=19 n-2
$$

thus

$$
13 m-5=19 n-2,
$$

from which

$$
m=19 p+9
$$

thus

$$
y=13(19 p+9)-5=247 p+112
$$

and

$$
x=11(247 p+112)+2=2717 p+1234
$$

Therefore, according to the condition,

$$
4000>2717 p+1234>1000
$$

This inequality can only be satisfied if $p$ is equal to 0 or 1. Therefore, the sought numbers are: 1234 and 3951.

(Géza Sasvári, Pécs.)

The problem was also solved by: Bayer B., Demeter J., Faith F., Filkorn J., Kerekes T., Krausz B., Krisztián Gy., Kürth A., Lukhaub Gy., Lupsa Gy., Messik G., Póka J., Sasvári J., Scharff J., Selényi M., Singer A.