We have an $n \times n$ square divided into unit squares. Each side of unit square is called unit segment. Some isoceles right triangles of hypotenuse $2$ are put on the square so all their vertices are also vertices of unit squares. For which $n$ it is possible that every unit segment belongs to exactly one triangle(unit segment belongs to a triangle even if it's on the border of the triangle)?
1. **Counting the Unit Segments:**
   - Consider an \( n \times n \) square grid. Each side of the unit square is called a unit segment.
   - The total number of unit segments in the grid can be calculated as follows:
     - Each row has \( n \) horizontal unit segments, and there are \( n+1 \) such rows (including the top and bottom boundaries).
     - Each column has \( n \) vertical unit segments, and there are \( n+1 \) such columns (including the left and right boundaries).
     - Therefore, the total number of unit segments is:
       \[
       2n(n+1)
       \]

2. **Properties of the Triangles:**
   - Each isosceles right triangle with hypotenuse \( 2 \) has exactly 3 unit segments (2 legs and 1 hypotenuse).
   - To cover all unit segments exactly once, the total number of unit segments must be divisible by 3.

3. **Divisibility Conditions:**
   - Since each triangle covers 3 unit segments, we need:
     \[
     3 \mid 2n(n+1)
     \]
   - This implies that \( n(n+1) \) must be divisible by 3. Since \( n \) and \( n+1 \) are consecutive integers, one of them must be divisible by 3. Therefore:
     \[
     3 \mid n \quad \text{or} \quad 3 \mid (n+1)
     \]

4. **Border Conditions:**
   - The unit segments on the border of the \( n \times n \) square are all contained on the diagonals of the triangles.
   - For the triangles to fit perfectly along the borders, \( n \) must be even. Therefore:
     \[
     2 \mid n
     \]

5. **Combining Conditions:**
   - Combining the conditions \( 3 \mid n(n+1) \) and \( 2 \mid n \), we get:
     \[
     n = 6k \quad \text{or} \quad n = 6k + 2 \quad \text{for some integer } k
     \]

6. **Constructing Examples:**
   - For \( n = 2 \):
     - We can place four triangles in a \( 2 \times 2 \) grid.
   - For \( n = 6 \):
     - We can place five \( 2 \times 2 \) blocks of triangles, one on each corner and one in the center, and fill the rest with triangles pointing in the same direction.
   - For the induction step from \( n \times n \) to \( (n+6) \times (n+6) \):
     - Take one \( n \times n \) grid and four blocks of triangles pointing in the same direction of size \( 2 \times (n+2) \).
     - Fill the remaining part with triangles to form a \( (n+6) \times (n+6) \) grid.

Therefore, we can construct examples for all \( n = 6k \) or \( n = 6k + 2 \).

The final answer is \( \boxed{ n = 6k } \) or \( n = 6k + 2 \).