18. (1994 Bulgarian Mathematical Olympiad) $n$ is a positive integer, $A$ is a set of subsets of the set $\{1,2, \cdots, n\}$, such that no element of $A$ contains another element of $A$. Find the maximum number of elements in $A$.
18. 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
$$
|A|=\sum_{k=1}^{n} f_{k} \text {. }
$$

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 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, thus we have
$$
\sum_{k=1}^{n} f_{k} \cdot k!(n-k)!\leqslant n!\text {. }
$$

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

When $A$ consists of all $\left[\frac{n}{2}\right]$-element subsets of $\{1,2, \cdots, n\}$, we achieve $|A|=C_{n}^{\frac{n}{2}}$.
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 $\frac{n}{2}$).