For which $n\ge 2$ is it possible to find $n$ pairwise non-similar triangles $A_1, A_2,\ldots , A_n$ such that each of them can be divided into $n$ pairwise non-similar triangles, each of them similar to one of $A_1,A_2 ,\ldots ,A_n$?
1. **Define the Problem and Notation:**
   We need to find for which \( n \geq 2 \) it is possible to find \( n \) pairwise non-similar triangles \( A_1, A_2, \ldots, A_n \) such that each of them can be divided into \( n \) pairwise non-similar triangles, each of them similar to one of \( A_1, A_2, \ldots, A_n \).

2. **Introduce the Triangle Notation:**
   Let \(\alpha + \beta < \frac{\pi}{2}\). We denote by \(\triangle(\alpha, \beta)\) a triangle with angles \(\alpha\) and \(\beta\). We write \(\triangle_1 = \triangle_2 + \triangle_3\) if \(\triangle_1\) can be cut into two triangles similar to \(\triangle_2\) and \(\triangle_3\) respectively.

3. **Construct the Triangles:**
   Consider small enough \(\alpha \ll 1\). Our triangles are \(\triangle(\alpha, \alpha), \triangle(\alpha, 2\alpha), \ldots, \triangle(\alpha, n\alpha)\).

4. **Cutting the Triangles:**
   Using one simple cut, we can get:
   \[
   \triangle(\alpha, k\alpha) = \triangle(\alpha, (k+1)\alpha) + \triangle(\alpha, k\alpha)
   \]
   and
   \[
   \triangle(\alpha, k\alpha) = \triangle(\alpha, (k-1)\alpha) + \triangle(\alpha, k\alpha)
   \]
   for each \( k \leq n \).

5. **Recursive Construction:**
   Using these formulas recursively, we get:
   \[
   \triangle(\alpha, k\alpha) = \triangle(\alpha, \alpha) + \triangle(\alpha, 2\alpha) + \cdots + \triangle(\alpha, n\alpha)
   \]
   for each \( 1 \leq k \leq n \).

6. **Conclusion:**
   Since we can construct such triangles for any \( n \geq 2 \), it is possible to find \( n \) pairwise non-similar triangles \( A_1, A_2, \ldots, A_n \) such that each of them can be divided into \( n \) pairwise non-similar triangles, each of them similar to one of \( A_1, A_2, \ldots, A_n \).

\(\blacksquare\)

The final answer is \( \boxed{ n \geq 2 } \).