Let $n$ be a positive integer $\geq 2$ . Consider a $n$ by $n$ grid with all entries $1$. Define an operation on a square to be changing the signs of all squares adjacent to it but not the sign of its own. Find all $n$ such that it is possible after a finite sequence of operations to reach a $n$ by $n$ grid with all entries $-1$
1. **Initial Setup and Definitions:**
   - We start with an \( n \times n \) grid where all entries are \( 1 \).
   - The operation defined changes the signs of all squares adjacent to a chosen square but does not change the sign of the chosen square itself.
   - We need to determine for which values of \( n \) it is possible to transform the entire grid to all entries being \( -1 \).

2. **Analyzing the Operation:**
   - Consider the effect of the operation on a single square. If we choose a square at position \((i, j)\), the squares affected are \((i-1, j)\), \((i+1, j)\), \((i, j-1)\), and \((i, j+1)\), provided these indices are within the grid boundaries.
   - Each operation changes the signs of exactly four squares (or fewer if the square is on the boundary or corner).

3. **Parity Consideration:**
   - Let's consider the parity (odd or even nature) of the number of \( 1 \)'s in the grid.
   - Initially, the grid has \( n^2 \) entries, all of which are \( 1 \). Thus, the number of \( 1 \)'s is \( n^2 \), which is even if \( n \) is even and odd if \( n \) is odd.

4. **Effect of Operations on Parity:**
   - Each operation changes the signs of an even number of squares (either 2 or 4). Therefore, each operation preserves the parity of the number of \( 1 \)'s in the grid.
   - If \( n \) is odd, \( n^2 \) is odd, and the number of \( 1 \)'s remains odd after any number of operations.
   - If \( n \) is even, \( n^2 \) is even, and the number of \( 1 \)'s remains even after any number of operations.

5. **Reaching the All \(-1\) Grid:**
   - To reach a grid where all entries are \( -1 \), the number of \( 1 \)'s must be zero.
   - For \( n \) odd, it is impossible to reach zero \( 1 \)'s because the number of \( 1 \)'s is always odd.
   - For \( n \) even, it is possible to reach zero \( 1 \)'s because the number of \( 1 \)'s can be reduced to zero, maintaining even parity throughout.

Conclusion:
- Therefore, it is possible to transform the grid to all \(-1\) entries if and only if \( n \) is even.

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