Consider a $ n \times n$ checkerboard with $ n > 1, n \in \mathbb{N}.$ How many possibilities are there to put $ 2n \minus{} 2$ identical pebbles on the checkerboard (each on a different field/place) such that no two pebbles are on the same checkerboard diagonal. Two pebbles are on the same checkerboard diagonal if the connection segment of the midpoints of the respective fields are parallel to one of the diagonals of the $ n \times n$ square.
1. **Define the Checkerboard and Diagonals:**
   Place the \( n \times n \) checkerboard on the coordinate plane such that the bottom left corner is at \( (1,1) \), the bottom right corner is at \( (n,1) \), and the top right corner is at \( (n,n) \).

2. **Identify the Diagonals:**
   Consider the following \( 2n - 2 \) diagonals:
   - Main diagonal: \( (1,1), (2,2), \ldots, (n,n) \)
   - Diagonals below the main diagonal:
     - \( (2,1), (1,2) \)
     - \( (3,1), (2,2), (1,3) \)
     - ...
     - \( (n-1,1), (n-2,2), \ldots, (2,n-1), (1,n) \)
     - \( (n,1), (n-1,2), \ldots, (2,n-1), (1,n) \)
   - Diagonals above the main diagonal:
     - \( (n,2), (n-1,3), \ldots, (3,n-1), (2,n) \)
     - ...
     - \( (n, n-2), (n-1,n-1), (n-2,n) \)
     - \( (n,n-1), (n-1,n) \)

3. **Pebble Placement Constraints:**
   These \( 2n - 2 \) diagonals span the entire board, so each diagonal must contain exactly one pebble. No two pebbles can be on the same diagonal.

4. **Determine Pebble Placement:**
   - For the main diagonal \( (1,1), (2,2), \ldots, (n,n) \), there are two places where the pebble can go: either on the first half or the second half of the diagonal.
   - For the diagonal \( (2,1), (1,2) \), there are two places where the pebble can go, and the placement of this pebble uniquely determines the placement of the pebble in the diagonal \( (n,n-1), (n-1,n) \).
   - For the diagonal \( (3,1), (2,2), (1,3) \), there are two possible placements since the \( (2,2) \) square is on the same diagonal as the \( (1,1) \) square which already contains a pebble. Each placement uniquely determines the placement of the pebble in the diagonal \( (n, n-2), (n-1,n-1), (n-2,n) \).

5. **Repeat the Process:**
   Clearly, we can repeat this process for all \( n \) diagonals containing squares in the bottom row, and the placement of pebbles in the other \( n-2 \) diagonals will be uniquely determined.

6. **Calculate the Number of Ways:**
   Since each of the \( n \) diagonals has 2 choices for placing the pebble, the total number of ways to place the pebbles is \( 2^n \).

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