$A_1,A_2,\cdots,A_8$ are fixed points on a circle. Determine the smallest positive integer $n$ such that among any $n$ triangles with these eight points as vertices, two of them will have a common side.
1. **Understanding the Problem:**
   We need to find the smallest positive integer \( n \) such that among any \( n \) triangles formed by the vertices \( A_1, A_2, \ldots, A_8 \) on a circle, at least two of the triangles share a common side.

2. **Initial Observation:**
   We start by noting that \( n > 8 \). This is because we can construct 8 triangles such that no two triangles share a common side. For example, consider the following set of triangles:
   \[
   \{A_1A_2A_3, A_3A_4A_5, A_5A_6A_7, A_7A_8A_1, A_2A_4A_7, A_4A_6A_1, A_6A_8A_3, A_8A_2A_5\}
   \]
   In this set, no two triangles share a common side.

3. **Using the Pigeonhole Principle:**
   If we have 9 triangles, we need to show that at least two of them must share a common side. Each triangle has 3 sides, so 9 triangles have a total of \( 9 \times 3 = 27 \) sides.

4. **Counting the Sides:**
   There are \(\binom{8}{2} = 28\) possible sides (edges) that can be formed from 8 points. However, each side can be used in at most one triangle without repetition. 

5. **Applying the Pigeonhole Principle:**
   By the Pigeonhole Principle, if we have 27 sides distributed among 28 possible sides, at least one side must be repeated. This means that at least two triangles must share a common side.

6. **Conclusion:**
   Therefore, the smallest positive integer \( n \) such that among any \( n \) triangles with these eight points as vertices, two of them will have a common side is \( n = 9 \).

The final answer is \( \boxed{ n = 9 } \).