The squares of an n*n chessboard (n >1) are filled with 1s and
-1s. A series of steps are performed. For each step, the number
in each square is replaced with the product of the numbers that
were in the squares adjacent. Find all values of n for which,
starting from an arbitrary collection of numbers, after finitely
many steps one obtains a board filled with 1s.
1. **Initial Setup and Definitions**:
   - Consider an \( n \times n \) chessboard where each square is filled with either \( 1 \) or \( -1 \).
   - Define a step where each number in a square is replaced by the product of the numbers in the adjacent squares (horizontally and vertically).

2. **Understanding the Transformation**:
   - Let \( a_{i,j} \) be the number in the square at position \((i, j)\).
   - After one step, the new value \( a'_{i,j} \) is given by:
     \[
     a'_{i,j} = a_{i-1,j} \cdot a_{i+1,j} \cdot a_{i,j-1} \cdot a_{i,j+1}
     \]
   - Note that the transformation depends on the parity (odd or even) of the number of \( -1 \)'s in the adjacent squares.

3. **Parity Analysis**:
   - If the number of \( -1 \)'s in the adjacent squares is even, the product is \( 1 \).
   - If the number of \( -1 \)'s in the adjacent squares is odd, the product is \( -1 \).

4. **Invariant Property**:
   - Consider the parity of the number of \( -1 \)'s on the entire board. This parity is invariant under the transformation.
   - If the total number of \( -1 \)'s is odd, it remains odd after each step.
   - If the total number of \( -1 \)'s is even, it remains even after each step.

5. **Final Configuration**:
   - For the board to be filled with \( 1 \)'s, the total number of \( -1 \)'s must be zero, which is an even number.
   - Therefore, the initial configuration must have an even number of \( -1 \)'s.

6. **Special Cases**:
   - For \( n = 2 \), consider the following configurations:
     - If the board starts with an even number of \( -1 \)'s, it can potentially reach a state of all \( 1 \)'s.
     - If the board starts with an odd number of \( -1 \)'s, it cannot reach a state of all \( 1 \)'s.

7. **General Case for \( n > 2 \)**:
   - The same parity argument applies. The board can only reach a state of all \( 1 \)'s if the initial configuration has an even number of \( -1 \)'s.

8. **Conclusion**:
   - The value of \( n \) does not affect the ability to reach a board filled with \( 1 \)'s. The key factor is the parity of the number of \( -1 \)'s in the initial configuration.

The final answer is \( \boxed{ n } \) can be any integer greater than 1, provided the initial configuration has an even number of \( -1 \)'s.