6) Let $X$ be a 56-element set. Find the smallest positive integer $n$, such that for any 15 subsets of $X$, if the union of any 7 of them has at least $n$ elements, then there must exist 3 of these 15 subsets whose intersection is non-empty. (Provided by Leng Gangsong)
The minimum value of $n$ is 41.
First, we prove that $n=41$ satisfies the conditions. We use proof by contradiction. Suppose there exist 15 subsets of $X$ such that the union of any 7 of them contains at least 41 elements, and the intersection of any 3 of them is empty. Since each element belongs to at most 2 subsets, we can assume each element belongs to exactly 2 subsets (otherwise, we can add some elements to some subsets, and the conditions still hold). By the pigeonhole principle, there must be a subset, say $A$, that contains at least $\left[\frac{2 \times 56}{15}\right]+1=8$ elements. Let the other 14 subsets be $A_{1}, A_{2}, \cdots, A_{14}$. Consider any 7 subsets that do not include $A$, each corresponding to 41 elements in $X$. All such 7-subset groups without $A$ together contain at least $41 C_{14}^{7}$ elements. On the other hand, for an element $a$, if $a \notin A$, then 2 of $A_{1}, A_{2}, \cdots, A_{14}$ contain $a$, so $a$ is counted $\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}$ times; if $a \in A$, then 1 of $A_{1}, A_{2}, \cdots, A_{14}$ contains $a$, so $a$ is counted $\mathrm{C}_{14}-\mathrm{C}_{13}^{7}$ times. Therefore,
$$
\begin{aligned}
41 C_{14}^{7} & \leqslant(56-|A|)\left(C_{14}^{7}-C_{12}^{7}\right)+|A|\left(C_{14}^{7}-C_{13}^{7}\right) \\
& =56\left(C_{14}^{7}-C_{12}^{7}\right)-|A|\left(C_{13}^{7}-C_{12}^{7}\right) \\
& \leqslant 56\left(C_{14}^{7}-C_{12}^{7}\right)-8\left(C_{13}^{7}-C_{12}^{7}\right),
\end{aligned}
$$

This leads to $196 \leqslant 195$, a contradiction.
Next, we prove that $n \geqslant 41$.
We use proof by contradiction. Suppose $n \leqslant 40$, and let $X=\{1,2, \cdots, 56\}$. Define
$$
\begin{array}{c}
A_{i}=\{i, i+7, i+14, i+21, i+28, i+35, i+42, i+49\}, \\
i=1,2, \cdots, 7, \\
B_{j}=\{j, j+8, j+16, j+24, j+32, j+40, j+48\}, \\
j=1,2, \cdots, 8 .
\end{array}
$$

Clearly, $\quad\left|A_{i}\right|=8(i=1,2, \cdots, 7)$,
$$
\begin{array}{c}
\left|A_{i} \cap A_{j}\right|=0(1 \leqslant i<j \leqslant 7), \\
\left|B_{j}\right|=7(j=1,2, \cdots, 8), \\
\left|B_{i} \cap B_{j}\right|=0(1 \leqslant i<j \leqslant 8), \\
\left|A_{i} \cap B_{j}\right|=1(1 \leqslant i \leqslant 7,1 \leqslant j \leqslant 8),
\end{array}
$$

Thus, for any 3 subsets, there must be 2 that are both $A_{i}$ or both $B_{j}$, and their intersection is empty.
For any 7 subsets
$$
A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{s}}, B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{t}}(s+t=7),
$$

we have
$$
\begin{aligned}
& \left|A_{i_{1}} \cup A_{i_{2}} \cup \cdots \cup A_{i_{s}} \cup B_{j_{1}} \cup B_{j_{2}} \cup \cdots \cup B_{j_{t}}\right| \\
= & \left|A_{i_{1}}\right|+\left|A_{i_{2}}\right|+\cdots+\left|A_{i_{s}}\right|+\left|B_{j_{1}}\right|+\left|B_{j_{2}}\right|+\cdots+\left|B_{j_{2}}\right|-s \\
= & 8 s+7 t-s t=8 s+7(7-s)-s(7-s) \\
= & (s-3)^{2}+40 \geqslant 40,
\end{aligned}
$$

Since the intersection of any 3 subsets is empty, it follows that $n \geqslant 41$.
In conclusion, the minimum value of $n$ is 41.