7.3. From the set of numbers $1,2,3,4, \ldots, 2021$, one number was removed, after which it turned out that the sum of the remaining numbers is divisible by 2022. Which number was removed?
Answer: 1011.

Solution. Let's write out the sum and perform grouping:

$1+2+3+\cdots+2019+2020+2021=(1+2021)+(2+2020)+$ $\cdots+(1012+1010)+1011$.

All the terms enclosed in parentheses are divisible by 2022. If we remove the last ungrouped term from the sum, the sum will be a multiple of 2022.

Comment. Answer only - 0 points.