A board $ n \times n$ ($n \ge 3$) is divided into $n^2$ unit squares. Integers from $O$ to $n$ included, are written down: one integer in each unit square, in such a way that the sums of integers in each $2\times 2$ square of the board are different. Find all $n$ for which such boards exist.
To solve the problem, we need to determine for which values of \( n \) it is possible to fill an \( n \times n \) board with integers from \( 0 \) to \( n \) such that the sums of integers in each \( 2 \times 2 \) square are different.

1. **Count the number of \( 2 \times 2 \) squares:**
   - An \( n \times n \) board contains \((n-1)^2\) different \( 2 \times 2 \) squares. This is because each \( 2 \times 2 \) square can start at any position \((i, j)\) where \( 1 \leq i, j \leq n-1 \).

2. **Determine the range of possible sums:**
   - Each \( 2 \times 2 \) square contains four integers, each ranging from \( 0 \) to \( n \). Therefore, the sum of the integers in any \( 2 \times 2 \) square ranges from \( 0 \) (if all four integers are \( 0 \)) to \( 4n \) (if all four integers are \( n \)).

3. **Compare the number of possible sums with the number of \( 2 \times 2 \) squares:**
   - The number of possible sums is \( 4n + 1 \) (from \( 0 \) to \( 4n \)).
   - For the sums to be different for each \( 2 \times 2 \) square, we need \( 4n + 1 \geq (n-1)^2 \).

4. **Solve the inequality \( 4n + 1 \geq (n-1)^2 \):**
   \[
   4n + 1 \geq (n-1)^2
   \]
   \[
   4n + 1 \geq n^2 - 2n + 1
   \]
   \[
   4n \geq n^2 - 2n
   \]
   \[
   0 \geq n^2 - 6n
   \]
   \[
   0 \geq n(n - 6)
   \]
   This inequality holds when \( n \leq 6 \).

5. **Verify the values of \( n \) from 3 to 6:**
   - For \( n = 3 \):
     \[
     4(3) + 1 = 13 \quad \text{and} \quad (3-1)^2 = 4
     \]
     Since \( 13 \geq 4 \), it is possible.
   - For \( n = 4 \):
     \[
     4(4) + 1 = 17 \quad \text{and} \quad (4-1)^2 = 9
     \]
     Since \( 17 \geq 9 \), it is possible.
   - For \( n = 5 \):
     \[
     4(5) + 1 = 21 \quad \text{and} \quad (5-1)^2 = 16
     \]
     Since \( 21 \geq 16 \), it is possible.
   - For \( n = 6 \):
     \[
     4(6) + 1 = 25 \quad \text{and} \quad (6-1)^2 = 25
     \]
     Since \( 25 \geq 25 \), it is possible.

Therefore, the values of \( n \) for which such boards exist are \( n = 3, 4, 5, 6 \).

The final answer is \( \boxed{n \in \{3, 4, 5, 6\}} \).