Russian Mathematical Olympiad

Suppose this is possible. Denote the given numbers by $a_1, a_2, \dots, a_{111}$ and let their sum be $S$. By the condition, for each index $k$, the numbers $a_k$ and $S - a_k$ have the same last digit. Hence, their difference $S - 2a_k$ is divisible by $10$. Therefore, for any $k$, the number $2a_k$ ends with the same digit as the sum $S$. This means that the difference between any two numbers $a_k$ is divisible by $5$. Thus, all $111$ distinct numbers $a_i$ must give the same remainder when divided by $5$; but among the numbers from $1$ to $500$, there are exactly $100$ numbers with any fixed remainder modulo $5$. Contradiction.