The 64th question: Let $S=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ be a non-empty family of subsets $U$ with the property: if $A \in U, A \subseteq B$, then $B \in U$; and let $S$ be a non-empty family of subsets $V$ with the property: if $A \in V, A \supseteq B$, then $B \in V$. Find the maximum possible value of $\frac{|U \cap V|}{|U| \cdot|V|}$.
The 64th question:
Solution: When $U=V=\{A \mid A \subseteq S\}$, it clearly satisfies the conditions, and $|U|=|V|=|U \cap V|=2^{n}$, at this time $\frac{|U \cap V|}{|U| \cdot|V|}=\frac{1}{2^{n}}$.
We now prove: $2^{n} \cdot|U \cap V| \leq|U| \cdot|V|$.
We use induction on $n$. When $n=1$, the conclusion is obviously true.
Assume the conclusion holds for $n-1$. Now consider the case for $n$.
Let the subfamily of subsets in $U$ that do not contain $a_{n}$ be $U_{1}$, and the subfamily of subsets in $U$ that contain $a_{n}$ be $U_{2}$. According to the conditions, we have $|U_{1}| \leq|U_{2}|$, and $|U_{1}|+|U_{2}|=|U|$.

Let the subfamily of subsets in $V$ that do not contain $a_{n}$ be $V_{1}$, and the subfamily of subsets in $V$ that contain $a_{n}$ be $V_{2}$. According to the conditions, we have $|V_{1}| \geq|U_{2}|$, and $|V_{1}|+|V_{2}|=|V|$.

At this point, $|U \cap V|=|U_{1} \cap V_{1}|+|U_{2} \cap V_{2}|$. According to the induction hypothesis, we have: $|U_{1} \cap V_{1}| \leq 2^{n-1}|U_{1}| \cdot|V_{1}|$, $|U_{2} \cap V_{2}| \leq 2^{n-1}|U_{2}| \cdot|V_{2}|$. Therefore, by the Chebyshev inequality, we have:
$|U \cap V|=|U_{1} \cap V_{1}|+|U_{2} \cap V_{2}| \leq \frac{|U_{1}| \cdot|V_{1}| + |U_{2}| \cdot|V_{2}|}{2^{n-1}} \leq \frac{(|U_{1}|+|U_{2}|)(|V_{1}|+|V_{2}|)}{2^{n}}=\frac{|U| \cdot|V|}{2^{n}}$
$\Rightarrow 2^{n} \cdot|U \cap V| \leq|U| \cdot|V|$
In summary, the maximum possible value of $\frac{|U \cap V|}{|U| \cdot|V|}$ is $\frac{1}{2^{n}}$.