Sofia and Viktor are playing the following game on a $2022 \times 2022$ board:
 - Firstly, Sofia covers the table completely by dominoes, no two are overlapping and all are inside the table;
 - Then Viktor without seeing the table, chooses a positive integer $n$;
 - After that Viktor looks at the table covered with dominoes, chooses and fixes $n$ of them;
 - Finally, Sofia removes the remaining dominoes that aren't fixed and tries to recover the table with dominoes differently from before.
If she achieves that, she wins, otherwise Viktor wins. What is the minimum number $n$ for which Viktor can always win, no matter the starting covering of dominoes.

[i]Proposed by Viktor Simjanoski[/i]
1. **Initial Setup and Strategy**:
   - Sofia covers the $2022 \times 2022$ board completely with dominoes. Each domino covers exactly two adjacent squares.
   - Viktor chooses a positive integer $n$ and then selects $n$ dominoes to fix in place.
   - Sofia then attempts to re-cover the board with the remaining dominoes in a different configuration.

2. **Minimum Number of Fixed Dominoes**:
   - We need to determine the minimum number $n$ such that Viktor can always ensure that Sofia cannot re-cover the board differently.
   - Consider the board as a grid of $2022 \times 2022$ squares. Each domino covers two adjacent squares, so there are $\frac{2022 \times 2022}{2} = 2022^2$ dominoes in total.

3. **Strategy for Viktor**:
   - If Sofia covers the board with $2 \times 2$ squares and places 2 dominoes in each square, then $n \geq 1011^2$ is necessary. Otherwise, Sofia can simply change the orientation of the two dominoes in some $2 \times 2$ square.
   - For $n = 1011^2$, Viktor's strategy is to fix the dominoes containing squares with odd coordinates. There are exactly $1011^2$ such dominoes.

4. **Proof by Contradiction**:
   - Suppose, for the sake of contradiction, that there is another way to cover the board with the remaining dominoes.
   - Connect two squares with a black edge if they are in one domino in the first cover, and with a white edge if they are in one domino in the second cover.
   - Ignore the dominoes that are common to both coverings (including the fixed ones). Each vertex (square) has degree 2, so the obtained graph is a union of alternating cycles with even length.

5. **Cycle Analysis**:
   - Consider the interior of some cycle on the board. Note that it is covered by dominoes, so there are an even number of squares there.
   - However, by induction on the number of "odd squares" in the interior, we can show that the number of squares in the interior is odd, leading to a contradiction.

6. **Conclusion**:
   - Therefore, the minimum number $n$ for which Viktor can always win is $1011^2$.

The final answer is $\boxed{1011^2}$