We see that the parity of the number of odd numbers on the board is invariant. A pair of even numbers reduces to an even difference, a pair of one odd number and one even number reduces to an odd difference, and a pair of two odd numbers reduces to an even difference, reducing the number of odd numbers by $2 \equiv 0 \mod 2$. We claim that a set of $N$ consecutive integers can be reduced to $0$, where $4|N$. This can be accomplished by repeatedly replacing consecutive numbers with their difference, $1$, until the board only consists of $1$s. Then, pairs of $1$s are combined. This necessarily results in a single $0$s, since there is always an even number of odd numbers in the original set, and so there is always an even amount of $1$s, so there will always be a pair to reduce to $0$. The first $2020$ numbers on the board constitute such a set of consecutive numbers, so reducing them to $0$ and then taking the difference of $2021$ and $0$ yields the desired result.

There are $1011$ odd numbers on the board, so $2020$ cannot be alone on the board, since that would imply that there is an even amount of odd numbers on the board.

We let the positions of $i$ be $a_i$. Do the operations $|a_{2i} - a_{2i-1}|$ for $i \in \{1,2,3,4,5,....,1010\}$. Now $1011$ numbers remain out of which $a_1 = a_2 = ...... = a_{1010} = 1$. Now do the operations $|a_{2i} - a_{2i-1}|$ for $i \in \{1,2,3,4,5,....,505\}$. Thus all numbers turn $0$ except $2021$.

Claim: The sum of all numbers $\pmod 2$ is constant. Proof: Say we pick $2$ numbers of the same parity, then their sum and difference is $0 \pmod 2$. Now if those $2$ numbers are of different parity then their sum and difference is $1 \pmod 2$. Thus whatever happens, it is constant $\pmod 2$. The sum of all the numbers in this case is $1 \pmod 2$ so if it can never result into the last number being even.