In each side of a regular polygon with $n$ sides, we choose a point different from the vertices and we obtain a new polygon of $n$ sides. For which values of $n$ can we obtain a polygon such that the internal angles are all equal but the polygon isn't regular?
1. **Define the problem and notation:**
   We are given a regular polygon with \( n \) sides. On each side of this polygon, we choose a point different from the vertices, forming a new polygon with \( n \) sides. We need to determine for which values of \( n \) we can obtain a polygon such that the internal angles are all equal but the polygon isn't regular.

2. **Case for even \( n \):**
   Let \( A_i \) be the vertices of the original regular polygon and \( P_i \) be the points chosen on \( A_iA_{i+1} \) (all indices are taken modulo \( n \)). Assume \( n \) is even. We can take the side length of the original polygon to be 1. Define the positions of \( P_i \) as follows:
   \[
   A_{2k+1}P_{2k+1} = \lambda \quad \text{and} \quad A_{2k}P_{2k} = 1 - \lambda
   \]
   for some \( \lambda \in (0, 1) \).

3. **Isosceles triangles and parallel sides:**
   Since \( A_iP_{i-1}P_i \) are all isosceles triangles, the sides \( P_iP_{i+1} \) are parallel to \( A_iA_{i+2} \). This ensures that all internal angles of the new polygon are equal. However, if \( \lambda \neq \frac{1}{2} \), the sides of the new polygon will have alternating lengths, making the polygon non-regular.

4. **Conclusion for even \( n \):**
   Therefore, for even \( n \), it is possible to construct a polygon with equal internal angles that is not regular.

5. **Case for odd \( n \):**
   Now consider \( n \) to be odd. Let \( P_1' = f_{\alpha}(P_1) \) (where \( 0 < \alpha < \frac{\pi}{n} \)) be the point on \( A_1A_2 \) such that \( \angle A_{i+1}P_iP_{i+1} = \alpha \) for \( i = 1, \ldots, n-1 \) and also \( \angle A_1P_nP_1' = \alpha \).

6. **Behavior of points \( P_i \):**
   If \( P_1 \) moves towards \( A_2 \), then \( P_1' \) moves towards \( A_1 \). By induction, \( P_{2k+1} \) moves towards \( A_{2k+2} \) and \( P_{2k} \) moves towards \( A_{2k} \). This implies that there can be at most one point \( P_1 \) such that \( P_1 = P_1' \).

7. **Symmetry and regularity:**
   If \( A_iP_i = \lambda \) with fixed \( \lambda \in (0, 1) \), we have a regular polygon (by symmetry), which also yields a solution for each \( \alpha \) in \( (0, \frac{\pi}{n}) \). These are all the possible values of \( \alpha \). Therefore, there can be no non-regular such polygon for odd \( n \).

8. **Conclusion for odd \( n \):**
   For odd \( n \), it is not possible to construct a polygon with equal internal angles that is not regular.

The final answer is all even \( \boxed{ n } \).