Problem 1524

Solution:
We may assume the set $A_{1}$ has maximal cardinality. Denote $A_{i} \cap A_{i+1} = B_{i}$, $i=1,2, \ldots, n$. Since $A_{n} \supset B_{n-1} \cup B_{n}$, then
$$\begin{aligned} \left|A_{n}\right| & \geq \left|B_{n-1} \cup B_{n}\right| = \left|B_{n-1}\right| + \left|B_{n}\right| - \left|B_{n-1} \cap B_{n}\right| \\ & > \frac{n-2}{n-1}\left|A_{n}\right| + \frac{n-2}{n-1}\left|A_{1}\right| - \left|B_{n-1} \cap B_{n}\right| \end{aligned}$$
Hence
$$\left|B_{n-1} \cap B_{n}\right| > \frac{n-2}{n-1}\left|A_{1}\right| - \frac{1}{n-1}\left|A_{n}\right| \geq \frac{n-3}{n-1}\left|A_{1}\right|$$
i.e., $\left|A_{n-1} \cap A_{n} \cap A_{1}\right| > \frac{n-3}{n-1}\left|A_{1}\right|$. Further, if $C = A_{n-1} \cap A_{n} \cap A_{1}$, then $A_{n-1} \supset C \cup B_{n-2}$ and
$$\begin{aligned} \left|A_{n-1}\right| & \geq \left|B_{n-2} \cup C\right| = \left|B_{n-2}\right| + |C| - \left|B_{n-2} \cap C\right| \\ & > \frac{n-2}{n-1}\left|A_{n-1}\right| + \frac{n-3}{n-1}\left|A_{1}\right| - \left|B_{n-2} \cap C\right| \end{aligned}$$
So $\left|B_{n-2} \cap C\right| > \frac{n-3}{n-1}\left|A_{1}\right| - \frac{1}{n-1}\left|A_{n-1}\right| \geq \frac{n-4}{n-1}\left|A_{1}\right|$, i.e.
$$\left|A_{n-2} \cap A_{n-1} \cap A_{n} \cap A_{1}\right| > \frac{n-4}{n-1}\left|A_{1}\right|$$
We get by induction that
$$\left|A_{n-k} \cap A_{n-k+1} \cap \cdots \cap A_{n-1} \cap A_{n} \cap A_{1}\right| > \frac{n-k-2}{n-1}\left|A_{1}\right|$$
for $k=1,2, \ldots, n-2$. In particular, $\left|A_{2} \cap A_{3} \cap \cdots \cap A_{n-1} \cap A_{n} \cap A_{1}\right| > 0$.