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 $A$.

[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 one of the $f_{k}$ $k$-element subsets $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$, 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 no two elements in $A$ are contained in 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 $\{1,2, \cdots, n\}$, we exactly achieve $|A|=C_{n}^{\left[\frac{n}{2}\right]}$.

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 $\frac{n}{2}$).