A cyclic $n$-gon is divided by non-intersecting (inside the $n$-gon) diagonals to $n-2$ triangles. Each of these triangles is similar to at least one of the remaining ones. For what $n$ this is possible?
1. **Initial Consideration**:
   - We are given a cyclic \( n \)-gon divided by non-intersecting diagonals into \( n-2 \) triangles.
   - Each triangle is similar to at least one of the remaining triangles.
   - We need to determine for which \( n \) this is possible.

2. **Case \( n = 3 \)**:
   - For \( n = 3 \), the polygon is a triangle itself, and there are no diagonals to draw. Hence, this case is trivial and does not satisfy the condition of having \( n-2 \) triangles.

3. **Case \( n = 4 \)**:
   - For \( n = 4 \), the polygon is a quadrilateral. By drawing one diagonal, we divide it into two triangles.
   - In a cyclic quadrilateral, the two triangles formed by the diagonal are similar if the quadrilateral is a rectangle or a square.
   - Therefore, \( n = 4 \) is a valid solution.

4. **Case \( n = 5 \)**:
   - For \( n = 5 \), the polygon is a pentagon. By drawing two non-intersecting diagonals, we divide it into three triangles.
   - These three triangles must be pairwise similar.
   - However, in a cyclic pentagon, it is impossible to draw two non-intersecting diagonals such that all three resulting triangles are similar. This is because at least two of the triangles will share a side or vertex, leading to a contradiction.
   - Therefore, \( n = 5 \) is not a valid solution.

5. **Case \( n = 2k \) for \( k \geq 2 \)**:
   - For \( n = 2k \), we consider a regular \( 2k \)-gon.
   - In a regular \( 2k \)-gon, drawing non-intersecting diagonals will divide it into \( 2k-2 \) triangles.
   - Each of these triangles will be isosceles and similar to each other due to the symmetry of the regular polygon.
   - Therefore, \( n = 2k \) for \( k \geq 2 \) is a valid solution.

6. **Case \( n = 2k + 3 \) for \( k \geq 2 \)**:
   - For \( n = 2k + 3 \), we consider a regular \( 2k \)-gon and add 3 points.
   - However, adding 3 points to a regular \( 2k \)-gon does not guarantee that the resulting triangles will be similar.
   - This construction does not necessarily lead to all triangles being similar, and thus, it is not a valid solution.

7. **Conclusion**:
   - From the above analysis, the valid solutions are \( n = 4 \) and \( n = 2k \) for \( k \geq 2 \).

The final answer is \( \boxed{ n = 4 } \) and \( n = 2k \) for \( k \geq 2 \).