Consider a latin square of size $n$. We are allowed to choose a $1 \times 1$ square in the table, and add $1$ to any number on the same row and column as the chosen square (the original square will be counted aswell) , or we can add $-1$ to all of them instead. Can we with doing finitly many operation , reach any latin square of size $n?$
1. **Lemma 1: For any 2 Rows and any Column, we can add 1 to the First Row and subtract 1 from the Second Row (except the 2 intersections of the Selected Column with 2 Rows) without changing the rest of the Square.**

   **Proof:**
   - Select one of the intersections and perform the operation by adding 1.
   - Select the other intersection and perform the operation by subtracting 1.
   - This ensures that the rest of the square remains unchanged.

2. **Lemma 2: For any 4 squares such that their centers create a rectangle, parallel to the sides of the square, we can change the numbers written in the 2 squares with the same row and add and subtract their difference to the other 2 squares.**

   **Proof:**
   - Select the 2 rows and one of the columns and perform the operation of Lemma 1, \(i-j\) times.
   - Select the 2 rows and the other column and perform the operation of Lemma 1, \(i-j\) times.
   - This will result in the desired transformation.

3. **Transforming the Latin Square:**
   - Change the numbers in the first row to match the first row of the target Latin square using Lemma 2, affecting only the last row.
   - Change the numbers in the second row to match the second row of the target Latin square using Lemma 2, affecting only the last row.
   - Continue this process for all rows except the last one.
   - The upper \(n \times (n-1)\) part of the square will now match the target Latin square.
   - Since Lemma 2 does not change the sum of each row and column, the numbers in the last row must match the numbers in the last row of the target Latin square.

4. **General Transformation:**
   - Let \(A\) and \(B\) be two Latin squares.
   - For every position \((x,y)\), calculate \(d = B[x,y] - A[x,y]\).
   - If \(d < 0\), add -1 for \(|d|\) times on the \(x\)-th row and \(y\)-th column.
   - If \(d > 0\), add +1 for \(d\) times.
   - This ensures that \(A\) is transformed into \(B\).

The final answer is that we can transform any Latin square \( \boxed{ A } \) to any Latin square \(B\) using the described operations.