Let's find four-digit numbers that, when divided by $2, 3, 5, 7$, and 11, respectively give remainders of $0, 1, 3, 5$, and 9.
We see that the remainder is always 2 less than the corresponding divisor. Therefore, if we subtract 2 from a number that is divisible by 2, 3, 5, 7, and 11, we get the number that meets the conditions. However, the given divisors are relatively prime, so their product is their least common multiple. This product is 2310. We get four-digit numbers if we multiply 2310 by 2, 3, and 4: 4620, 6930, 9240. Subtracting 2 from these numbers, we get the desired numbers:

$2308, 4618, 6928, 9238$.

(Pál Ligeti, Budapest.)

The problem was also solved by: Aczél F., Baranyó E., Bartók I., Bayer B., Beck P., Blau A., Bogdán G., Demjén E., Deutsch I., Ehrenstein P., Enyedi B., Freund E., Haar A., Hirschfeld Gy., Kelemen M., Kertész F., Kertész G., König D., Kürti I., Lázár L., Pilczer P., Pintér M., Pivnyik I., Póka Gy., Riesz K., Riesz M., Schlesinger A., Schmidl I., Schwarz Gy., Szmodics H., Szávay Z., Ungár B., Weisz P., Wohlstein S.