1. **Define the Graph $G$**:
- Let $G$ be a graph where each vertex represents one of the subsets $A_1, A_2, \ldots, A_{101}$.
- Draw an edge between two vertices if the corresponding subsets have a common element.

2. **Assume $G$ is Triangle-Free**:
- Suppose $G$ does not contain any triangles, i.e., there are no three vertices such that each pair of them is connected by an edge.

3. **Degree of Vertices in $G$**:
- We will show that every vertex in $G$ has a degree of at most 50.
- Assume for contradiction that there exists a vertex $v$ with degree at least 51. Let $v$ be adjacent to vertices $v_1, v_2, \ldots, v_{51}$.

4. Union of Subsets:
- Consider the sets corresponding to $v_1, v_2, \ldots, v_{51}$. The union of these 51 sets must have more than $\frac{50}{51}n$ elements.
- By the Pigeonhole Principle, at least one of these sets, say $A_{v_{51}}$, must have more than $\frac{n}{51}$ elements.

5. Contradiction:
- Now consider the sets corresponding to $v_1, v_2, \ldots, v_{50}$. The union of these 50 sets must also have more than $\frac{50}{51}n$ elements.
- Since $A_{v_{51}}$ has more than $\frac{n}{51}$ elements, there must be some overlap with the union of $A_{v_1}, A_{v_2}, \ldots, A_{v_{50}}$.
- Therefore, at least one of $v_1, v_2, \ldots, v_{50}$ must be adjacent to $v_{51}$, forming a triangle $vv_1v_{51}$, which contradicts our assumption that $G$ is triangle-free.

6. Existence of Triangles:
- Since $G$ cannot be triangle-free, there must be at least one triangle in $G$.
- In fact, there are at least 18 triangles in $G$. Suppose there are at most 17 triangles: $a_1b_1c_1, a_2b_2c_2, \ldots, a_kb_kc_k$ where $k \leq 17$.
- This leaves at least $101 - 3 \cdot 17 = 50$ vertices not in these triangles. Call these vertices "free vertices".

7. Degree of Free Vertices:
- A free vertex $v$ can have at most degree 50 (otherwise, it would form a triangle with some other vertices).
- Therefore, at least 50 vertices are not connected to $v$, implying $|v| < \frac{n}{51}$.

8. Union of Free Vertices:
- The union of the sets corresponding to these free vertices has less than $\frac{50}{51}n$ elements, which contradicts the given condition that the union of any 50 subsets has more than $\frac{50}{51}n$ elements.

Thus, we have shown that there must exist at least three subsets among the 101 subsets such that any two of them have a common element.

$\blacksquare$