Suppose $X$ is a set with $|X| = 56$. Find the minimum value of $n$, so that for any 15 subsets of $X$, if the cardinality of the union of any 7 of them is greater or equal to $n$, then there exists 3 of them whose intersection is nonempty.
1. **Define the problem and assumptions:**
   We are given a set \( X \) with \( |X| = 56 \). We need to find the minimum value of \( n \) such that for any 15 subsets of \( X \), if the cardinality of the union of any 7 of them is greater than or equal to \( n \), then there exists 3 of them whose intersection is nonempty.

2. **Prove that \( n = 41 \) is sufficient:**
   Assume we have subsets \( A_1, A_2, \ldots, A_{15} \subseteq X \) such that \( |A_{i_1} \cup A_{i_2} \cup \cdots \cup A_{i_7}| \geq 41 \) for any 7 subsets. We need to show that there exist 3 subsets whose intersection is nonempty.

3. **Contradiction approach:**
   Suppose, for contradiction, that every element of \( X \) occurs in at most 2 sets. If there is some \( x \in X \) which belongs to a single \( A_i \), we can add this \( x \) to some other \( A_j \) without affecting the property that the union of any 7 is at least 41. Thus, we can assume that every element of \( X \) occurs in exactly 2 sets.

4. **Graph construction:**
   Construct a graph \( G \) with vertices \( 1, 2, \ldots, 15 \) where vertices \( i \) and \( j \) are connected if \( A_i \cap A_j \neq \emptyset \). Since each element of \( X \) is in exactly 2 sets, \( G \) has exactly 56 edges.

5. **Edge counting:**
   For any 7 vertices, the number of edges with at least one endpoint among these 7 vertices is at least 41. Therefore, for any 8 vertices, the number of edges with both endpoints among them is at most \( 56 - 41 = 15 \).

6. **Degree analysis:**
   Consider a vertex \( v \) with degree \( \deg(v) \). The number of edges among the remaining 14 vertices is \( 56 - \deg(v) \leq 15 \). This implies:
   \[
   56 - \deg(v) \leq 15 \implies \deg(v) \geq 41
   \]
   However, this is incorrect. The correct inequality should be:
   \[
   56 - \deg(v) \leq 15 \implies \deg(v) \geq 41
   \]
   This is a contradiction because the maximum degree in a graph with 15 vertices and 56 edges cannot be this high.

7. **Counterexample for \( n = 40 \):**
   To show that \( n = 40 \) is not sufficient, consider a bipartite graph \( K_{8,7} \). This graph has 56 edges, and the union of any 7 subsets can be made to have a cardinality of 40, but there will be no 3 subsets with a nonempty intersection.

The final answer is \( \boxed{41} \).