Let $n$ be a positive integer. A panel of dimenisions $2n\times2n$ is divided in $4n^2$ squares with dimensions $1\times1$. What is the highest possible number of diagonals that can be drawn in $1\times1$ squares, such that each two diagonals have no common points. 
1. **Understanding the Problem:**
   We have a \(2n \times 2n\) panel divided into \(4n^2\) squares of dimensions \(1 \times 1\). We need to find the maximum number of diagonals that can be drawn in these \(1 \times 1\) squares such that no two diagonals share a common point.

2. **Counting Vertices:**
   The \(2n \times 2n\) panel has \((2n+1) \times (2n+1)\) vertices. This is because each side of the panel has \(2n+1\) vertices (including the boundaries).

3. **Drawing Diagonals:**
   Each \(1 \times 1\) square can have one of its two diagonals drawn. We need to ensure that no two diagonals share a common vertex.

4. **Strategy for Drawing Diagonals:**
   To avoid any two diagonals sharing a common vertex, we can use a checkerboard pattern. In a checkerboard pattern, we can draw diagonals in every alternate square.

5. **Checkerboard Pattern:**
   - For a \(2n \times 2n\) grid, there are \(2n \times 2n = 4n^2\) squares.
   - In a checkerboard pattern, half of these squares will have diagonals, and the other half will not.
   - Therefore, the number of squares with diagonals is \(\frac{4n^2}{2} = 2n^2\).

6. **Verification:**
   - In a checkerboard pattern, no two diagonals will share a common vertex because each diagonal is separated by at least one square without a diagonal.
   - This ensures that the maximum number of diagonals is achieved without any two diagonals intersecting.

7. **Example for \(n=4\):**
   - For \(n=4\), the panel is \(8 \times 8\).
   - Using the checkerboard pattern, we can draw diagonals in \(2 \times 4^2 = 32\) squares.

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