The numbers from $1$ to $n^2$ are written in order in a grid of $n \times n$, one number in each square, in such a way that the first row contains the numbers from $1$ to $n$ from left to right; the second row contains the numbers $n + 1$ to $2n$ from left to right, and so on and so forth. An allowed move on the grid consists in choosing any two adjacent squares (i.e. two squares that share a side), and add (or subtract) the same integer to both of the numbers that appear on those squares. 

Find all values of $n$ for which it is possible to make every squares to display $0$ after making any number of moves as necessary and, for those cases in which it is possible, find the minimum number of moves that are necessary to do this. 
1. **Understanding the Problem:**
   - We have an \( n \times n \) grid filled with numbers from \( 1 \) to \( n^2 \) in a sequential manner.
   - We are allowed to perform moves where we choose two adjacent squares and add or subtract the same integer to both numbers in those squares.
   - The goal is to determine for which values of \( n \) it is possible to make all squares display \( 0 \) and find the minimum number of moves required.

2. **Initial Observations:**
   - Each number in the grid is unique and ranges from \( 1 \) to \( n^2 \).
   - A move affects two adjacent squares, and we need to make all numbers zero.

3. **Feasibility for \( n \):**
   - We need to determine if it is possible to make all numbers zero for any \( n \).
   - If \( n \) is odd, it is impossible to make all numbers zero because the sum of all numbers from \( 1 \) to \( n^2 \) is \(\frac{n^2(n^2 + 1)}{2}\), which is odd, and we cannot split this sum into equal parts using moves that affect pairs of numbers.

4. **Proof for Even \( n \):**
   - For even \( n \), we can use a strategy to make all numbers zero.
   - Consider dividing the \( n \times n \) grid into \( 2 \times 2 \) subgrids. Each \( 2 \times 2 \) subgrid can be manipulated to make all its numbers zero using three moves.

5. **Minimum Number of Moves:**
   - Each \( 2 \times 2 \) subgrid requires three moves to make all its numbers zero.
   - There are \(\frac{n^2}{4}\) such \( 2 \times 2 \) subgrids in an \( n \times n \) grid.
   - Therefore, the total number of moves required is \( 3 \times \frac{n^2}{4} = \frac{3n^2}{4} \).

6. **Detailed Calculation:**
   - For each \( 2 \times 2 \) subgrid, label the numbers as follows:
     \[
     \begin{array}{|c|c|}
     \hline
     a & b \\
     \hline
     c & d \\
     \hline
     \end{array}
     \]
   - Perform the following moves:
     1. Subtract \( a \) from both \( a \) and \( b \) (affects \( a \) and \( b \)).
     2. Subtract \( c \) from both \( c \) and \( d \) (affects \( c \) and \( d \)).
     3. Subtract \( b \) from both \( b \) and \( d \) (affects \( b \) and \( d \)).
   - After these moves, all numbers in the \( 2 \times 2 \) subgrid will be zero.

7. **Graph Theory Approach:**
   - Consider a graph \( G \) where each vertex represents a square in the grid, and edges represent moves between adjacent squares.
   - Each connected component in \( G \) must have at least 4 vertices because each move affects two squares, and we need at least two moves to make a square zero.
   - The number of edges \( e \) in \( G \) satisfies:
     \[
     e \geq \frac{3n^2}{4}
     \]
   - This confirms that the minimum number of moves required is \(\frac{3n^2}{4}\).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{3n^2}{4} } \) moves for even \( n \).