Example 6 Let $A=\{1,2, \cdots, n\}$, where $n$ is a positive integer. A subset of $A$ is called connected if it contains only one positive integer from $A$ or only several consecutive positive integers from $A$. Determine the maximum positive integer $k$, such that there exist $k$ different subsets $A_{1}, A_{2}, \cdots, A_{k}$ of $A$, such that the intersection of any two different subsets is connected.
(2000 Taiwan Olympiad Problem)
Let $A_{1}, A_{2}, \cdots, A_{k}$ be the subsets that satisfy the problem's requirements, and let $\min A_{i}, \max A_{i}$ denote the smallest and largest numbers in the subset $A_{i}$, respectively. Let $m=\max _{1 \leqslant i \leqslant k}\left\{\min A_{i}\right\}$, and suppose $m=\min A_{i_{0}}$.

By the definition of $m$, the smallest number in each subset $A_{i}$ is less than or equal to $m$, and the largest number in each subset $A_{i}$ is greater than or equal to $m$ (since if $\max A_{i}<m$, and $\min A_{i_{0}}=m$, then $A_{i} \cap A_{i_{0}}=\varnothing$, which contradicts the fact that $A_{i} \cap A_{i_{0}}$ is connected). Since $k$ pairs of numbers $\left(\min A_{i}, \max A_{i}\right)$ each equal one of the $m(n+1-m)$ pairs $(r, s)$, where $1 \leqslant r \leqslant m \leqslant s \leqslant n$.

For each such pair of numbers $(r, s)(1 \leqslant r \leqslant m \leqslant s \leqslant n)$, we prove that there is at most one $A_{i}$ such that $\left(\min A_{i}, \max A_{i}\right)=(r, s)$. Indeed, if there are two such subsets $A_{i}$ and $A_{j}(i \neq j)$ both corresponding to the same $(r, s)$, then $A_{i} \cap A_{j}$ is a connected set containing $r$ and $s$, so $A_{i}=A_{j}=\{r, r+1, r+2, \cdots, s-1, s\}$, which contradicts $A_{i} \neq A_{j}$. Therefore, $k$ is at most
$$
m(n+1-m)=-\left(m-\frac{n+1}{2}\right)^{2}+\frac{(n+1)^{2}}{4} .
$$

Thus, $k \leqslant\left[\frac{(n+1)^{2}}{4}\right]$.
The equality can be achieved by taking all connected subsets of $A$ that contain $\left[\frac{n+1}{2}\right]$, and the number of such subsets is exactly
$$
\left[\frac{n+1}{2}\right]\left(n+1-\left[\frac{n+1}{2}\right]\right)=\left[\frac{(n+1)^{2}}{4}\right] \text { . }
$$

In summary, the maximum value of $k$ is $\left[\frac{(n+1)^{2}}{4}\right]$.