$11 \cdot 90$ For any positive integer $k$, try to find the smallest positive integer $f(k)$, such that there exist 5 sets $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$, satisfying the following conditions:
(1) $\left|S_{i}\right|=k, i=1,2,3,4,5$;
(2) $S_{i} \cap S_{i+1}=\varnothing\left(S_{6}=S_{1}\right), i=1,2,3,4,5$;
(3) $\left|\bigcup_{i=1}^{5} S_{i}\right|=f(k)$.

Also, ask what the result is when the number of sets is a positive integer $n(n \geqslant 3)$.
[Solution] We only solve the second question.

From (2), we know that each of the $f(k)$ numbers appears at most $\left[\frac{n}{2}\right]$ times in the $n$ sets, and the $n$ sets contain a total of $n k$ numbers, so we have
$$
f(k) \geqslant \frac{n k}{\left[\frac{n}{2}\right]}.
$$

Since $f(k)$ is a positive integer, we have
$$
f(k) \geqslant\left[\frac{n k+\left[\frac{n}{2}\right]-1}{\left[\frac{n}{2}\right]}\right].
$$

We will prove that the equality in the above formula holds by construction. When $n$ is even, $f(k) = 2k$, and the conclusion is obviously true. Now, let $n$ be odd, $n=2 m+1 (m \geqslant 1)$, then (1) becomes
$$
f(k)=2 k+1+\left[\frac{k-1}{m}\right].
$$

Let $k=p m+q$, where $0 \leqslant q < m$.
$$
\begin{array}{l}
\left\{\begin{array}{l}
f(k)=2 k+1+\left[\frac{k-1}{m}\right] \\
k=p m+q, \quad 0 \leqslant q < m
\end{array}\right.
\end{array}\right.
$$

Let the $2 k+p+1$ numbers be $a_{1}, a_{2}, \cdots, a_{2 k+p+1}$, and construct a sequence with $n k$ terms as follows: it consists of $m$ segments, the first $q$ segments each have $2 k+p+1$ terms, which are $a_{1}, a_{2}, \cdots, a_{2 k+p+1}$, and the last $m-q$ segments each have $2 k+p$ terms, which are $a_{1}, a_{2}, \cdots, a_{2 k+p}$. Note that when $q=0$, only $a_{2 k+p}$ different numbers are used. Let the sequence thus formed be
$$
b_{1}, b_{2}, \cdots, b_{(2 k+p+1) q+(2 k+p)(m-q)}.
$$

Let
$$
S_{i}=\left\{b_{k i-k+j} \mid j=1,2, \cdots, k\right\}, i=1,2, \cdots, n,
$$

then this set of sets satisfies condition (2), so (3) holds.
For $i=1,2, \cdots, n-1$, it is obvious that $S_{i} \cap S_{i+1}=\varnothing$, so we only need to verify that $S_{n} \cap S_{1}=\varnothing$. From (4), we have
$$
S_{n}=\left\{b_{2 m k+j} \mid j=1,2, \cdots, k\right\}.
$$

Since $2 m k=(2 k+p+1) q+(2 k+p)(m-q-1)+k+p$, we have $b_{2 m k+j}=a_{k+p+j}, j=1,2, \cdots, k$, thus
$$
S_{n}=\left\{a_{k+p+j} \mid j=1,2, \cdots, k\right\}.
$$

It is clear that $S_{n} \cap S_{1}=\varnothing$. Therefore, we have proved that the equality in (1) holds.