How many ways are there to remove an $11\times11$ square from a $2011\times2011$ square so that the remaining part can be tiled with dominoes ($1\times 2$ rectangles)?
1. **Coloring the Grid:**
   - Consider the $2011 \times 2011$ square grid. Color the grid in a checkerboard pattern, starting with the top-left cell as black. This means that each black cell is adjacent to white cells and vice versa.
   - In a $2011 \times 2011$ grid, there are $2011^2$ cells. Since $2011$ is odd, there will be one more black cell than white cells. Specifically, there will be $\left\lceil \frac{2011^2}{2} \right\rceil$ black cells and $\left\lfloor \frac{2011^2}{2} \right\rfloor$ white cells.

2. **Removing an $11 \times 11$ Square:**
   - When we remove an $11 \times 11$ square, we need to consider the color of the top-left cell of this smaller square.
   - An $11 \times 11$ square has $11^2 = 121$ cells. If the top-left cell of this $11 \times 11$ square is black, then there will be $\left\lceil \frac{121}{2} \right\rceil = 61$ black cells and $\left\lfloor \frac{121}{2} \right\rfloor = 60$ white cells in the $11 \times 11$ square.
   - Conversely, if the top-left cell of the $11 \times 11$ square is white, then there will be $60$ black cells and $61$ white cells in the $11 \times 11$ square.

3. **Remaining Grid After Removal:**
   - If the $11 \times 11$ square removed has a black top-left cell, the remaining $2000 \times 2000$ grid will have an equal number of black and white cells, making it possible to tile the remaining grid with $1 \times 2$ dominoes.
   - If the $11 \times 11$ square removed has a white top-left cell, the remaining grid will have two more black cells than white cells, making it impossible to tile the remaining grid with $1 \times 2$ dominoes.

4. **Counting Valid Removals:**
   - The $2011 \times 2011$ grid can be divided into $2001 \times 2001$ possible positions for the top-left corner of the $11 \times 11$ square.
   - Out of these, half will have a black top-left cell and half will have a white top-left cell. Since $2011$ is odd, there will be one extra position with a black top-left cell.
   - Therefore, the number of valid ways to remove an $11 \times 11$ square such that the remaining part can be tiled with dominoes is:
     \[
     \left\lceil \frac{2001^2}{2} \right\rceil = \left\lceil \frac{2001 \times 2001}{2} \right\rceil = \left\lceil 2001 \times 1000.5 \right\rceil = 2001 \times 1001 = 2,002,001
     \]

The final answer is $\boxed{2,002,001}$.