, 2020

(a) First we observe that:
At every movement $K$, the numbers $\alpha + \beta$ and $(\alpha - \beta)^{2020}$ are congruent modulo $2$, that is, both are even or both are odd. Therefore, after the substitution of $\alpha, \beta$ with the number $(\alpha - \beta)^{2020}$ the parity of the sum of the numbers on the blackboard remains invariant. Since

$$S_{2030} = \frac{2030 \cdot 2031}{2} = 1015 \cdot 2031$$
is odd, it follows that after the last movement the number remaining on the blackboard is odd and therefore the answer in the first question is negative.

(b) Let after performing the movement enough number of times we have on the blackboard two numbers $x, y$. Then the only case in order no one of them is not a power of $2$, is not to have taken part in any performed movement. This can happen only with the numbers $1$ and $2030$. Since $1$ is $1^{2020}$, it is enough to check $2030$. If $x = a^{2020}$ and $y = 2030$, then we must have $(a^{2020} - 2030)^{2020} = 2021^{2020}$. For $a = 1$ there is no solution and for $a \ge 2$ we get $a^{2020} = 4051$, impossible.

In any other case we will have from the procedure that $x = (a_1 - b_1)^{2020} = m^{2020}$ and $y = (a_2 - b_2)^{2020} = n^{2020}$, for some $m$ and $n$. By performing the last movement, remains on the blackboard the number $(m^{2020} - n^{2020})^{2020}$. We have to check if it is possible to be valid the following equality:

$$(m^{2020} - n^{2020})^{2020} = 2021^{2020}.$$

If $m = n$, it is not valid. We suppose wlog $m > n$. Then we must check if

$$m^{2020} - n^{2020} = 2021. \qquad (1)$$

The least possible value of $m^{2020} - n^{2020}$ is $1$, for $m = 1, n = 0$. From (1) we have that $m > n$, and hence $m \ge n + 1$. Therefore $m^{2020} - n^{2020} \ge (n+1)^{2020} - n^{2020}$. By developing the last expression we observe that all coefficients of $n$ have positive sign and so it is increasing. Hence $m^{2020} - n^{2020} \ge (n+1)^{2020} - n^{2020} \ge 2^{2020} - 1 > 2^{11} - 1 = 2047$ and therefore it cannot be equal to $2021$.