3. $n$ 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 $A$.
(1994 Bulgarian Olympiad Problem)
3. 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, 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 that exactly form the subset $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ are $k!(n-k)!$ in number. Since any two elements in $A$ are distinct, the permutations obtained by the above method for all elements in $A$ are necessarily distinct. Therefore, we have $\sum_{k=1}^{n} f_{k} \cdot k!(n- \sum_{k=1}^{n} \frac{f_{k}}{C_{n}^{k}}=C_{n}^{\left\lfloor\frac{n}{2}\right\rfloor} \cdot \frac{1}{n!} \sum_{k=1}^{n} f_{k} \cdot k!(n-k)!\leqslant C_{n}^{\left\{\frac{n}{2}\right]}$. 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 $\frac{n}{2}$).