 Determine all integer $n > 3$ for which a regular $n$-gon can be divided into equal triangles by several (possibly intersecting) diagonals.

(B.Frenkin)
1. **Identify the problem**: We need to determine all integer \( n > 3 \) for which a regular \( n \)-gon can be divided into equal triangles by several (possibly intersecting) diagonals.

2. **Consider even \( n \)**: 
   - For even \( n \), we can draw all the main diagonals from one vertex to all other non-adjacent vertices. This will divide the \( n \)-gon into \( n \) equal triangles.
   - For example, in a regular hexagon (\( n = 6 \)), drawing diagonals from one vertex to all other non-adjacent vertices divides the hexagon into 6 equal triangles.

3. **Consider odd \( n \)**:
   - Label the vertices \( V_1, V_2, V_3, \ldots, V_n \) in clockwise order.
   - Consider the shortest diagonal that is drawn, say \( V_1 V_{i+1} \) where \( i \le \frac{n-1}{2} \).
   - Notice that the triangle which contains the side \( V_1V_2 \) must also have a side contained in \( V_1 V_{i+1} \). An analogous thing holds for \( V_iV_{i+1} \).
   - If the triangle which contains \( V_1 V_2 \) is not isosceles, then the diagonals of shortest length must pair up the vertices (e.g., \( (V_1, V_{i+1}) \) is one pair). This is absurd.

4. **Analyze the diagonals**:
   - All diagonals of the form \( V_j V_{j+i} \) are drawn, for \( 1 \le j \le n \) (indices modulo \( n \)).
   - If \( i = 2 \), then it's easy to see that this fails.
   - If \( i = \frac{n-1}{2} \), it's easy to see that there is an \( n \)-gon in the center of the polygon, which is absurd.
   - Else, suppose that \( 2 < i < \frac{n-1}{2} \). Note that the diagonals \( V_1 V_{i+1} \) and \( V_2 V_{i+2} \) intersect at an angle of \( \frac{2 \pi}{n} \). Therefore, one angle of the triangle which "tiles" the polygon is at most \( \frac{2\pi}{n} \).

5. **Check for specific \( i \)**:
   - By looking at the triangle containing \( V_1 V_2 \), this implies that \( i = 3 \).
   - But then, it's easily checked that the triangle bounded by \( V_1 V_4, V_2 V_5, V_3V_6 \) is smaller than that bounded by \( V_1V_4, V_{n-1}V_2, V_1 V_2 \). This is absurd.

6. **Conclusion**:
   - Since the above analysis shows that odd \( n \) does not work, we conclude that only even \( n \) can be divided into equal triangles by several (possibly intersecting) diagonals.

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