$11 \cdot 80$ Let $n \in N$, find the largest $k \in N$, such that there exist $k$ distinct subsets in an $n$-element set, where the intersection of any two subsets is non-empty.
［Solution］Let $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, and consider all subsets of $S$ that contain $a_{1}$. Clearly, the number of such subsets is equal to the number of all subsets of $\left\{a_{2}, a_{3}, \cdots, a_{n}\right\}$, which is $2^{n-1}$. The intersection of any two such subsets is obviously non-empty, so we have $k \geqslant 2^{n-1}$.

On the other hand, for any group of $2^{n-1}+1$ subsets of $S$, we can divide all $2^{n}$ subsets of $S$ into $2^{n-1}$ pairs, where each subset $M \subset S$ is paired with its complement $S-M$ relative to $S$. By the pigeonhole principle, among the $2^{n-1}+1$ subsets, there must be two subsets that belong to the same pair among the above $2^{n-1}$ pairs, thus their intersection is an empty set.
In summary, the maximum natural number $k=2^{n-1}$.