We claim that $n$ is at most $\boxed{4}$. For $n = 4$, let $A_1, A_2, A_3, A_4$ be a chain of squares such that $A_1$ is connected to $A_2$, $A_2$ is connected to $A_3$, $A_3$ is connected to $A_4$, and $A_4$ is connected to $A_1$. Let $B_1 = A_3, B_2 = A_4, B_3 = A_1,$ and $B_4 = A_2$. This satisfies the property. Now let $n \ge 5$. Call a number in $\{1, 2, \ldots, n\}$ westward if the $x$-coordinates of all points in $A_i$ is less than that if all points on $B_i$. Define eastward, northward, and southward analogously. By pigeonhole principle, there are at least $2$ numbers of one category. WLOG assume there are $2$ distinct westward numbers $i$ and $j$. Let $(x, y)$ be a point in $A_i \cap B_j$ and $(x', y')$ be a point in $A_j \cap B_i$. But since $A_i$ is entirely west of $B_i$, $x < x'$, and since $A_j$ is entirely west of $B_j$, $x' < x$. Contradiction. Therefore, $n \le 4$.