20.1.3 Let $X$ be a 56-element set. Find the smallest 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.
Parse the minimum value of $n$ as 41.
First, prove that $n=41$ meets the conditions. Use proof by contradiction. Assume 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, add or remove some elements in some subsets, and the conditions still hold). By the pigeonhole principle, there must be a subset, say $A$, containing 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 7-element subsets that do not include $A$ together correspond to at least $41 \mathrm{C}_{14}^{7}$ 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 $\mathrm{C}_{14}^{7}-\mathrm{C}_{13}^{7}$ times. Thus,
$$
\begin{aligned}
41 \mathrm{C}_{14}^{7} & \leqslant(56-|A|)\left(\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}\right)+|A|\left(\mathrm{C}_{14}^{7}-\mathrm{C}_{13}^{7}\right) \\
& =56\left(\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}\right)-|A|\left(\mathrm{C}_{13}^{7}-\mathrm{C}_{12}^{7}\right) \\
& \leqslant 56\left(\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}\right)-8\left(\mathrm{C}_{13}^{7}-\mathrm{C}_{12}^{7}\right),
\end{aligned}
$$

This leads to $196 \leqslant 195$, a contradiction.
Next, prove that $n \geqslant 41$.
Use proof by contradiction. Assume $n \leqslant 40$, 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,
$$
\begin{array}{c}
\left|A_{i} \cap A_{j}\right|=0(1 \leqslant i<j \leqslant 7), \\
\left|B_{j}\right|=7(j=1,2, \cdots, 8),\left|B_{i} \cap B_{j}\right|=0(1 \leqslant i<j \leqslant 8), \\
\left|A_{i} \cap B_{j}\right|=1(1 \leqslant i \leqslant 7,1 \leqslant j \leqslant 8),
\end{array}
$$

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
$$
A_{t_{1}}, A_{i_{2}}, \cdots, A_{i_{1}}, B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{t}}(s+t=7),
$$

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

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