Four, (50 points) Find the maximum value of $m$ such that from the subsets of an $n$-element set, $m$ different subsets $A_{1}, A_{2}, \cdots, A_{m}$ can be selected, satisfying $\left(A_{i} \cap A_{k}\right) \subseteq A_{j}$ for all $1 \leqslant i<j<k \leqslant m$.


Let's assume this $n$-element set is $M=\{1,2, \cdots, n\}$.
For any $i \in M$, let the subsets of $M$ that contain $i$ be $A_{i_{1}}, A_{i_{i}}, \cdots, A_{i_{i}}\left(i_{1}<i_{2}<\cdots<i_{t}\right)$.

Given that $\left(A_{i} \cap A_{k}\right) \subseteq A_{j}$ for all $1 \leqslant i<j<k \leqslant m$, it follows that $i_{1}, i_{2}, \cdots, i_{1}$ are consecutive positive integers.
Consider the adjacent set pairs in $A_{1}, A_{2}, \cdots, A_{m}$:
$\left(A_{1}, A_{2}\right),\left(A_{2}, A_{3}\right), \cdots,\left(A_{m-1}, A_{m}\right)$.
When $i$ belongs to exactly one of the sets $A_{k}$ or $A_{k+1}$, we call $\left(A_{k}, A_{k+1}\right)$ an "i-separating set pair".

For any $1 \leqslant j \leqslant m-1$, $\left(A_{j}, A_{j+1}\right)$ must be separated by some element, otherwise $A_{j}=A_{j+1}$, which is a contradiction.

Furthermore, any element in set $M$ can separate at most two set pairs, thus
$$
m-1 \leqslant 2 n \Rightarrow m \leqslant 2 n+1 .
$$

When $m=2 n+1$, it must be that $A_{1}=A_{2 m+1}=\varnothing$, which is a contradiction. Therefore, $m \leqslant 2 n$.
$$
\begin{array}{l}
\text { Also, } A_{1}=\varnothing, A_{2}=\{1\}, A_{3}=\{1,2\}, \cdots, \\
A_{n+1}=\{1,2, \cdots, n\}, A_{n+2}=\{2,3, \cdots, n\}, \\
\cdots, A_{2 n}=\{n\},
\end{array}
$$

In this case, the selected subsets satisfy the problem's conditions, and $m=2 n$.
In summary, the maximum value of $m$ is $2 n$.
(Song Hongjun, Fuyang High School, Zhejiang, 311400)