Each cell of $2024\times 2024$ table colored either red, bule, or white. We place a red pawn on each red cell, and a blue pawn on each blue cell. Moreover, for each white cell that shares an edge or vertex with one or more blue cells, place a blue pawn on it. Suppose that for any $2\times2$ square, the number of red pawns and blue pawns placed within it is equal and either one or two. Find the maximum possible value for the number of squares painted white.
1. **Consider a \(4 \times 4\) table:**
   - We need to ensure that each \(2 \times 2\) sub-square within the \(4 \times 4\) table has an equal number of red and blue pawns, either 1 or 2.
   - We start by noting that there must be at least one blue cell in the \(4 \times 4\) table. This is because if there were no blue cells, the \(2 \times 2\) sub-squares would not satisfy the condition of having an equal number of red and blue pawns.

2. **Divide the \(4 \times 4\) table into four \(2 \times 2\) sub-squares:**
   - The sub-square containing the blue cell must have 2 red cells to balance the number of pawns.
   - The remaining three \(2 \times 2\) sub-squares must each contain at least one red cell to ensure the balance of pawns.

3. **Calculate the minimum number of red cells:**
   - The sub-square with the blue cell has 2 red cells.
   - The other three sub-squares each have at least 1 red cell.
   - Therefore, the total number of red cells is at least \(2 + 3 \times 1 = 5\).

4. **Determine the number of white cells:**
   - The total number of cells in the \(4 \times 4\) table is \(16\).
   - Subtracting the number of red cells (5) and the blue cell (1), we get \(16 - 5 - 1 = 10\) white cells.

5. **Verify the configuration:**
   - The following \(4 \times 4\) table configuration satisfies the conditions and has 10 white cells:
     \[
     \begin{array}{|c|c|c|c|}
     \hline
     r & w & w & w \\
     \hline
     w & w & r & w \\
     \hline
     w & r & b & r \\
     \hline
     w & w & r & w \\
     \hline
     \end{array}
     \]

6. **Generalize to a \(4n \times 4n\) table:**
   - Divide the \(4n \times 4n\) table into \(n^2\) non-overlapping \(4 \times 4\) sub-squares.
   - Each \(4 \times 4\) sub-square can have at most 10 white cells, as shown above.
   - Therefore, the total number of white cells in the \(4n \times 4n\) table is \(10n^2\).

7. **Optimality of \(10n^2\):**
   - Each \(4 \times 4\) sub-square must contain at least one blue cell to satisfy the conditions.
   - As shown, each \(4 \times 4\) sub-square must have at least 5 red cells.
   - Thus, the number of white cells in each \(4 \times 4\) sub-square cannot exceed 10.
   - Hence, the maximum number of white cells in the \(4n \times 4n\) table is \(10n^2\).

\(\blacksquare\)

The final answer is \( \boxed{ 10n^2 } \)