Austrian Mathematical Olympiad

(a) Let us first choose the following $1010$ pairs of numbers: $(1,2);(3,4);\dots ;(2019,2020)$. The absolute value of the difference within each of these pairs is $1$. After applying the operation for each of these pairs, the number $2021$ and $1010$ times the number $1$ remain on the blackboard. Now we execute the given operation $505$ times with pairs of the form $(1,1)$. Then the number $2021$ and $505$ times the number $0$ remain on the blackboard. As $2021-0=2021$ and $0-0=0$, we end up with $2021$ as the final number on the board after additional $505$ operations, regardless of the pairs we pick at each step.

(b) We prove a more general statement: The final remaining number on the blackboard cannot be even. As $$a-b\equiv a+b(\mathrm{mod}2),$$ we obtain that the parity of the sum of all numbers on the board is an invariant throughout the game. At the beginning, the sum of the numbers on the blackboard is $$\frac{2021\cdot 2022}{2}=2021\cdot 1011,$$ an odd number. Therefore, the final number on the board must be odd as well. In particular, $2020$ cannot be the final number on the blackboard.