4. A three-digit number, in which there are no zeros, is written on the board. The sum of all different numbers that can be obtained by rearranging the digits of the written number is 2775. What number could have been written on the board
Answer: $889,988,898,997,799,979$.

Solution. The number on the board could not have been written with three identical digits, as then the number on the board would be 2775, which is not a three-digit number. But it also could not have been written with three different digits, because then each digit in the units place would have appeared in two numbers, and the sum would have been even. Therefore, the original number is written with two different digits $a$ and $b$, leading to the equation $\overline{a a b}+\overline{a b a}+\overline{b a a}=2775$. When adding these numbers, we get $2 a+b$ units, $2 a+b$ tens, and $2 a+b$ hundreds. Therefore, $(2 a+b)(100+10+1)=2775$. Hence, $2 a+b=25$. Therefore, the numbers on the board could have been numbers with a sum of digits 25, that is, 889 or 997, and, accordingly, all numbers obtained from them by permuting the digits.

Comment. Proven that the number cannot be written with three different digits - 1 point. Proven that $2 a+b=25$ - 5 points. Correct solution found only one set of digits (889 or 997) - 6 points. Correct solution missed some permutations of the numbers 889 and 997 - 7 points. Without explanation, the case where the number is written with three identical digits is discarded - deduct 1 point. For potentially useful ideas and approaches in the absence of a solution - 2-3 points.