Example 8 Let $X$ be a set with 56 elements. Find the smallest positive integer $n$, such that for any 15 subsets of $X$, if the union of any seven of these subsets has at least $n$ elements, then there must exist three of these 15 subsets whose intersection is non-empty. ${ }^{[2]}$
(2006, China Mathematical Olympiad)
First, we prove by contradiction that \( n = 41 \) satisfies the requirement.
Assume there exist 15 subsets \( A_{1}, A_{2}, \cdots, A_{15} \), such that the union of any seven subsets has at least 41 elements, but the intersection of any three subsets is empty.

At this point, let \( Y = \{A_{1}, A_{2}, \cdots, A_{15}\} \), then \( X + Y \) forms a bipartite graph, where \( x \in X \) is adjacent to \( A_{j} \in Y \) if and only if \( x \) is an element of \( A_{j} \).

Since the intersection of any three subsets is empty, each element in \( X \) appears in at most two subsets, i.e., the degree of each vertex in \( X \) is less than or equal to 2.

Without loss of generality, assume each vertex in \( X \) has a degree of 2 (if a vertex has a degree \( d < 2 \), we can arbitrarily add \( 2 - d \) neighbors).

Thus, the number of edges in the graph is \( 2 \times 56 = 112 \), and the graph has exactly 56 (vertices in \( X \)) angles.

For any seven vertices in \( Y \), their number of neighbors is at least 41, which means the number of angles related to these seven vertices is at least 41. Thus, for the remaining eight vertices, the number of angles formed by any two of them is at most \( 56 - 41 = 15 \).

Next, fix 14 vertices in \( Y \), among which there are \( \binom{14}{8} \) "eight-point groups," and each eight-point group has at most 15 angles, while each angle appears in \( \binom{12}{6} \) eight-point groups. Therefore, the number of angles formed by these 14 vertices is less than or equal to \( 15 \times \frac{\binom{14}{8}}{\binom{12}{6}} < 49 \).

This indicates that the number of angles related to the remaining vertex in \( Y \) is greater than 7, i.e., the degree of this vertex is at least 8.

Thus, the degree of any vertex in \( Y \) is at least 8, and the number of edges in the graph is at least \( 8 \times 15 = 120 \). This contradicts the fact that the number of edges is 112.
In conclusion, \( n = 41 \) satisfies the given requirements.

Next, we show that when \( n = 40 \), there exist 15 subsets \( A_{1}, A_{2}, \cdots, A_{15} \), such that the union of any seven subsets has at least 40 elements, but the intersection of any three subsets is empty.

In fact, consider such a bipartite graph \( X + Y \), where \( Y = \{A_{1}, A_{2}, \cdots, A_{15}\} \), and for each \( 1 \leq i \leq 7 \) and \( 8 \leq j \leq 15 \), there is exactly one vertex in \( X \) adjacent to both \( A_{i} \) and \( A_{j} \) (i.e., it forms an angle with them).

It is easy to verify that any eight vertices in \( Y \) form at most \( x(8 - x) \leq 16 \) angles, thus, the number of angles related to the remaining seven vertices is at least 40, i.e., their number of neighbors is at least 40.
In conclusion, the smallest positive integer that satisfies the requirements is 41.