Example 40 Find the maximum positive integer $n$, such that $I=\{1,2, \cdots, n\}$ can be partitioned into two disjoint subsets $A$ and $B$ satisfying that no three numbers in the same subset form an arithmetic progression.
When $n=8$, let $A=\{1,4,5,8\}, B=\{2,3,6,7\}$, then any three numbers in the same subset of $A$ or $B$ do not form an arithmetic sequence.

When $n \geqslant 9$, if $I$ can be decomposed into two disjoint subsets $A$ and $B$ such that any 3 numbers in the same subset do not form an arithmetic sequence, then there are only the following three cases:
(1) 1 and 2 belong to the same subset, 3 belongs to the other subset. Without loss of generality, assume $1,2 \in A, 3 \in B$.
(a) If $4 \in A$, then $6,7 \in B$, thus $5,8 \in A$, hence the three numbers $2,5,8$ in $A$ form an arithmetic sequence, a contradiction.
(b) If $4 \in B$, then $5 \in A$, thus $8,9 \in B$, which implies $6,7 \in A$, hence the three numbers $5,6,7$ in $A$ form an arithmetic sequence, a contradiction.
(2) 1 and 3 belong to the same subset, 2 belongs to the other subset. Without loss of generality, assume $1,3 \in A, 2 \in B$, thus $5 \in B$, $8 \in A$.
(a) If $4 \in A$, then $7 \in B$, thus $9 \in A, 6 \in B$, hence the three numbers $5,6,7$ in $B$ form an arithmetic sequence, a contradiction.
(b) If $4 \in B$, then $6 \in A$, thus $7,9 \in B$, hence the three numbers $5,7,9$ in $B$ form an arithmetic sequence, a contradiction.
(3) 2 and 3 belong to the same subset, 1 belongs to the other subset. Without loss of generality, assume $1 \in A, 2,3 \in B$, thus $4 \in A$, $7 \in B$, which implies $5 \in A, 9 \in B$, thus $6 \in A$, hence the three numbers $4,5,6$ in $A$ form an arithmetic sequence, a contradiction. Therefore, when $n \geqslant 9$, there does not exist a decomposition method that meets the requirements of the problem, so the maximum value of $n$ is 8.