Four, (50 points) Let $X$ be a 56-element set. Find the smallest positive integer $n$, such that for any 15 subsets of set $X$, if the union of any seven of them has at least $n$ elements, then there must exist three subsets among these 15 such that their intersection is non-empty.
Four, construct 15 subsets
$$
\begin{array}{l}
A_{i}=\{i, i+1, \cdots, i+7\}(i=1,2, \cdots, 7) \text {, } \\
B_{j}=\{j, j+8, j+16, j+24, j+32 \text {, } \\
j+40, j+48\}(j=1,2, \cdots, 8) . \\
\end{array}
$$

Then $\left|A_{i}\right|=8(i=1,2, \cdots, 7)$,
$$
\begin{array}{l}
\left|B_{j}\right|=7(j=1,2, \cdots, 8), \\
\left|A_{i} \cap A_{j}\right|=\left|B_{i} \cap B_{j}\right|=0(i \neq j), \\
\left|A_{i} \cap B_{j}\right|=1(1 \leqslant i \leqslant 7,1 \leqslant j \leqslant 8) .
\end{array}
$$

Thus, in these 15 subsets, any three must include two from the $A$ group or two from the $B$ group, and the intersection of any three is an empty set.
Now, analyze the number of elements in any seven of these subsets.
Take any seven subsets $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{1}}$ and $B_{j_{1}}$, $B_{j_{2}}, \cdots, B_{j_{i}}$, where $s+t=7$. Then
$$
\begin{array}{l}
\left|A_{i_{1}} \cup A_{i_{2}} \cup \cdots \cup A_{i_{s}} \cup B_{j_{1}} \cup B_{j_{2}} \cup \cdots \cup B_{j_{1}}\right| \\
=8 s+7 t-s t=8 s+7(7-s)-s(7-s) \\
=(s-3)^{2}+40 \geqslant 40 .
\end{array}
$$

Therefore, the positive integer $n \geqslant 41$.
Next, use proof by contradiction to show that $n=41$ satisfies the condition.
Assume there exists a set $X$ with 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, meaning each element belongs to at most two subsets.
Consider two cases.
(1) Each element of set $X$ belongs to exactly two of the subsets $A_{1}, A_{2}$, $\cdots, A_{15}$.

By the pigeonhole principle, there must be a subset (let's say $A_{15}$) that contains at least $\left[\frac{2 \times 56}{15}\right]+1=8$ elements (where $[x]$ denotes the greatest integer less than or equal to the real number $x$). The union of all seven-element subsets formed by $A_{1}, A_{2}, \cdots, A_{14}$ must include at least $41 \mathrm{C}_{14}^{7}$ elements.

On the other hand, for any element $a$, if $a \notin A_{15}$, then only two of the subsets $A_{1}, A_{2}, \cdots, A_{14}$ contain $a$, so $a$ is counted $\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}$ times; if $a \in A_{15}$, then only one of the subsets $A_{1}, A_{2}, \cdots, A_{14}$ contains $a$, so $a$ is counted $\mathrm{C}_{14}^{7}-\mathrm{C}_{13}^{7}$ times.
Thus, $41 \mathrm{C}_{14}^{7}$
$$
\begin{array}{l}
\leqslant\left(56-\left|A_{15}\right|\right)\left(\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}\right)+\left|A_{15}\right|\left(\mathrm{C}_{14}^{7}-\mathrm{C}_{13}^{7}\right) \\
=56\left(\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}\right)-\left|A_{15}\right|\left(\mathrm{C}_{13}^{7}-\mathrm{C}_{12}^{7}\right) \\
\leqslant 56\left(\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}\right)-8\left(\mathrm{C}_{13}^{7}-\mathrm{C}_{12}^{7}\right),
\end{array}
$$

which implies $196 \leqslant 195$, a contradiction.
(2) Some elements of set $X$ may belong to at most one of the subsets $A_{1}, A_{2}, \cdots, A_{15}$.

In the subsets that do not contain these elements, add one element to each until every element of set $X$ belongs to exactly two of the subsets $A_{1}, A_{2}, \cdots, A_{15}$. Thus, the intersection of any three of the modified subsets $A_{1}, A_{2}, \cdots, A_{15}$ remains empty. This case is now reduced to (1), which is also a contradiction.

In summary, for any 15 subsets of set $X$, if the union of any seven of them has at least 41 elements, then there must exist three subsets whose intersection is non-empty.
Therefore, the smallest positive integer $n$ that satisfies the condition is $41$.