2・111 $n$ is a positive integer, $A$ is a set of subsets of the set $\{1,2, \cdots, n\}$, such that no element in $A$ contains another element in $A$, find the maximum number of elements in all elements of $A$.

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2・111 $n$ is a positive integer, $A$ is a set of subsets of the set $\{1,2, \cdots, n\}$, such that no element in $A$ contains another element in $A$, find the maximum number of elements in all elements of $A$.
[Solution] Consider the permutations of $n$ elements $1,2, \cdots, n$. Clearly, the number of such permutations is $n!$.

Let $A$ contain $f_{k}$ $k$-element subsets as elements, where $f_{k}(k=1,2, \cdots, n)$ are non-negative integers. Let $|A|$ denote the number of elements in $A$, then
$$|A|=\sum_{k=1}^{n} f_{k}$$

When $f_{k}>0$, for any $k$-element subset $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ among the $f_{k}$ $k$-element subsets, in the permutations of $1,2, \cdots, n$, the permutations where the first $k$ elements exactly form the subset $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ are $k!(n-k)!$ in number.

Since any two elements in $A$ do not contain each other, the permutations obtained by the above method for all elements in $A$ are necessarily distinct. Therefore, we have
$$\sum_{k=1}^{n} f_{k} \cdot k!(n-k)!\leqslant n!$$

Noting that, for a fixed positive integer $n$, $C_{n}^{k}$ reaches its maximum value when $k=\left[\frac{n}{2}\right]$, we have
$$\begin{aligned}
|A| & =\sum_{k=1}^{n} f_{k} \leqslant C_{n}^{\left[\frac{n}{2}\right]} \sum_{k=1}^{n} \frac{f_{k}}{C_{n}^{k}} \\
& =C_{n}^{\left[\frac{n}{2}\right]} \cdot \frac{1}{n!} \sum_{k=1}^{n} f_{k} \cdot k!(n-k)! \\
& \leqslant C_{n}^{\left[\frac{n}{2}\right]}
\end{aligned}$$

When $A$ is the set of all $\left[\frac{n}{2}\right]$-element subsets of $\{1,2, \cdots, n\}$, it exactly reaches $|A|=C_{n}^{\left[\frac{n}{2}\right]}$.

In summary, the maximum number of elements in $A$ is $C_{n}^{\left[\frac{n}{2}\right]}$ (where $\left[\frac{n}{2}\right]$ denotes the greatest integer not exceeding $\frac{n}{2}$).