Example 6 Let $n>1$ be a positive integer. Divide the set $S$ of all diagonals of a convex $(4n-1)$-gon into $k$ subsets $S_{1}, S_{2}, \cdots, S_{k}$, such that for every pair of distinct indices $i, j$, some diagonals in subset $S_{i}$ intersect with some diagonals in subset $S_{j}$ (intersection means intersecting at an interior point of the segments). Find the maximum possible value of $k$. ${ }^{[7]}$
Solve $k_{\max }=(n-1)(4 n-1)$.
First, prove by contradiction:
$k \leqslant(n-1)(4 n-1)$.
Otherwise, the number of all diagonals of a $(4 n-1)$-gon in the plane is $2(n-1)(4 n-1)$, i.e.,
$$
|S|=2(n-1)(4 n-1) \text {. }
$$

Therefore, there must exist an $i$ such that $\left|S_{i}\right|=1$, meaning there is a set containing only one diagonal, denoted as $l$.

Assume diagonal $l$ has $m$ and $4 n-3-m$ vertices on its two sides. To ensure that other sets contain diagonals intersecting with $l$, there must be at least one diagonal in other sets connecting the points on both sides of $l$.
$$
\begin{array}{l}
\text { Hence, if } k>(n-1)(4 n-1), \text { then } A_{k}=A_{k-4 n+1}, \text { where } i=1, 2, \cdots, 4 n-1, j=2,3, \cdots, n.
\end{array}
$$
Next, verify that this partition satisfies the problem's requirements.
(1) It is easy to see that there are $(n-1)(4 n-1)$ mutually disjoint subsets $S_{(i, j)}$, and $S=\cup S_{(i, j)}$, satisfying the conditions.
(2) Verify that for any $(i, j) \neq(m, n)$, there is always a diagonal in $S_{(i, j)}$ that intersects with a diagonal in $S_{(m, n)}$.

By rotational symmetry, assume $i=0$, and the two vertices of a diagonal without internal intersection points with the diagonals in set $S_{(0, j)}$ must be taken from one of the following sets:
$$
\begin{array}{l}
\left\{A_{0}, A_{1}, A_{2}, \cdots, A_{j-1}\right\}, \\
\left\{A_{j}, A_{j+1}, A_{j+2}, \cdots, A_{2 n}\right\}, \\
\left\{A_{2 n}, A_{2 n+1}, A_{2 n+2}, \cdots, A_{4 n}\right\} .
\end{array}
$$

Since none of the above three sets contain more than $2 n$ consecutive vertices, no set can contain the vertices of two diagonals in $S_{(m, n)}$. Thus, it implies that there is always a diagonal in $S_{(m, n)}$ that intersects with a diagonal in $S_{(0, j)}$ at an internal point.