Let $k$ be a positive integer. Arnaldo and Bernaldo play a game in a table $2020\times 2020$, initially all the cells are empty. In each round a player chooses a empty cell and put one red token or one blue token, Arnaldo wins if in some moment, there are $k$ consecutive cells in the same row or column with tokens of same color, if all the cells have a token and there aren't $k$ consecutive cells(row or column) with same color, then Bernaldo wins. If the players play alternately and Arnaldo goes first, determine for which values of $k$, Arnaldo has the winning strategy.
To determine for which values of \( k \) Arnaldo has a winning strategy, we need to analyze the game on a \( 2020 \times 2020 \) table. The key is to understand the conditions under which Arnaldo can force a win by creating \( k \) consecutive cells in the same row or column with tokens of the same color.

1. **Understanding the Game Setup:**
   - The table is \( 2020 \times 2020 \).
   - Arnaldo wins if there are \( k \) consecutive cells in the same row or column with tokens of the same color.
   - Bernaldo wins if all cells are filled and there are no \( k \) consecutive cells with the same color in any row or column.

2. **Generalizing the Problem:**
   - For a general \( 2n \times 2n \) table, we need to determine the critical value of \( k \) for which Arnaldo has a winning strategy.
   - The conjecture is that Arnaldo wins for \( k \leq n+1 \) and Bernaldo wins for \( k \geq n+2 \).

3. **Bernaldo's Winning Strategy for \( k \geq n+2 \):**
   - Consider the \( 2n \times 2n \) table.
   - Pair the cells in such a way that each pair is assigned a unique number.
   - For example, in a \( 8 \times 8 \) table, the pairing can be done as shown in the problem statement.
   - When Arnaldo places a token in a cell with a certain number, Bernaldo places a token of the opposite color in the cell with the same number.
   - This ensures that any sequence of \( n+2 \) consecutive cells will have at least two cells with different colors, preventing Arnaldo from winning.

4. **Arnaldo's Winning Strategy for \( k \leq n+1 \):**
   - To prove that Arnaldo has a winning strategy for \( k = n+1 \), we can use induction.
   - Base Case: For \( n = 1 \), the table is \( 2 \times 2 \), and Arnaldo can easily win by placing two tokens of the same color in a row or column.
   - Inductive Step: Assume Arnaldo can win for a \( 2(n-1) \times 2(n-1) \) table with \( k = n \). We need to show that Arnaldo can win for a \( 2n \times 2n \) table with \( k = n+1 \).
   - Arnaldo can use a strategy that forces Bernaldo into a position where he cannot prevent \( n+1 \) consecutive cells of the same color.

5. **Conclusion:**
   - By the pairing strategy, Bernaldo can prevent Arnaldo from winning for \( k \geq n+2 \).
   - By induction, Arnaldo can force a win for \( k \leq n+1 \).

Thus, for a \( 2020 \times 2020 \) table, \( n = 1010 \), and the critical value of \( k \) is \( 1011 \).

The final answer is \( \boxed{ k \leq 1011 } \).