Task 3. On the board, all such natural numbers from 3 to 223 inclusive are written, which when divided by 4 leave a remainder of 3. Every minute, Borya erases any two of the written numbers and instead writes their sum, decreased by 2. In the end, only one number remains on the board. What can it be?
Answer: 6218.

Solution. Initially, the total number of written numbers is $224: 4=56$, and their sum is $(3+223) \cdot 56: 2=6328$. Every minute, the number of written numbers decreases by 1, and their sum decreases by 2. Therefore, one number will remain on the board after 55 minutes and will be equal to $6328-55 \cdot 2=6218$.