Given is a positive integer $n$. There are $2n$ mutually non-attacking rooks placed on a grid $2n \times 2n$. The grid is splitted into two connected parts, symmetric with respect to the center of the grid. What is the largest number of rooks that could lie in the same part?
1. **Understanding the Problem:**
   We are given a $2n \times 2n$ grid with $2n$ mutually non-attacking rooks. The grid is split into two connected parts, symmetric with respect to the center. We need to determine the largest number of rooks that can lie in the same part.

2. **Constructing an Example:**
   To show that $2n-1$ rooks can be placed in one part, consider the following construction:
   - Split the grid along the main diagonal from the top left to the bottom right.
   - Assign the upper left half of the diagonal to the upper right piece and the lower right half to the lower left piece.
   - This results in a symmetric decomposition into two parts.
   - Place $2n-1$ rooks on the diagonal from the upper right to the main diagonal and the last rook on the lower left corner.

3. **Proving the Upper Bound:**
   To show that it is not possible to have all $2n$ rooks in one part, consider the following argument:
   - Any decomposition of the grid into two connected parts must miss at least one column or one row.
   - If a part of piece A is in both the left and right columns, and a part of piece B is in both the upper and lower rows, then there must be a path inside the pieces going from left to right and top to bottom.
   - Since both pieces are connected, these paths must cross, which is a contradiction.

4. **Conclusion:**
   Therefore, the largest number of rooks that can lie in the same part is $2n-1$.

The final answer is $\boxed{2n-1}$.