1. Consider rectangles in the plane with sides parallel to the coordinate axes (both length and width are greater than 0), and call such a rectangle a "box". If two boxes have a common point (including common points on the interior or boundary of the boxes), then the two boxes are said to "intersect". Find the largest positive integer $n$, such that there exist $n$ boxes $B_{1}, B_{2}, \cdots, B_{n}$, satisfying $B_{i}$ intersects $B_{j}$ if and only if $i \neq j \pm 1(\bmod n)$.
1. The maximum value that satisfies the condition is 6.
An example is shown in Figure 1.

Below is the proof: 6 is the maximum value.

Assume the boxes $B_{1}$, $B_{2}, \cdots, B_{n}$ satisfy the condition. Let the closed intervals corresponding to the projections of $B_{k}$ on the $x$-axis and $y$-axis be $I_{k}$ and $J_{k}$ respectively $(1 \leqslant k \leqslant n)$.

If $B_{i}$ and $B_{j}$ intersect, and the common point is $(x, y)$, then $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$. Therefore, $I_{i} \cap I_{j}$ and $J_{i} \cap J_{j}$ are non-empty. Conversely, if there exist real numbers $x$ and $y$ such that $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$, then $(x, y)$ is a common point of $B_{i}$ and $B_{j}$. Thus, $B_{i}$ and $B_{j}$ do not intersect if and only if at least one of the intervals $I_{i}, I_{j}$ and $J_{i}, J_{j}$ do not intersect, i.e., either $I_{i} \cap I_{j}=\varnothing$ or $J_{i} \cap J_{j}=\varnothing$.

For convenience, if the indices of the boxes differ by 1 modulo $n$, then the two boxes or intervals are called "adjacent." Other cases are called non-adjacent.

For each $k=1,2, \cdots, n$, the adjacent boxes $B_{k}$ and $B_{k+1}$ do not intersect, so $\left(I_{k}, I_{k+1}\right)$ or $\left(J_{k}, J_{k+1}\right)(1 \leqslant k \leqslant n)$ is a pair of non-intersecting intervals. Therefore, among $\left(I_{1}, I_{2}\right), \cdots,\left(I_{n-1}, I_{n}\right),\left(I_{n}, I_{1}\right)$; $\left(J_{1}, J_{2}\right), \cdots,\left(J_{n-1}, J_{n}\right),\left(J_{n}, J_{1}\right)$, there are at least $n$ pairs of non-intersecting intervals.

Since any two non-adjacent boxes intersect, the corresponding intervals on both coordinate axes also intersect.

Proposition: Let $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$ be $n$ closed intervals on the number line, and any two non-adjacent intervals intersect. Then the number of pairs of non-intersecting intervals $\left(\Delta_{k}, \Delta_{k+1}\right)(k=1,2, \cdots, n)$ is at most 3.

Proof of the proposition: Let $\Delta_{k}=\left[a_{k}, b_{k}\right](1 \leqslant k \leqslant n)$, $\alpha=\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be the rightmost left endpoint of $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$, and $\beta=\min \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ be the leftmost right endpoint of $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$. Without loss of generality, assume $\alpha=a_{2}$.

If $\alpha \leqslant \beta$, then for all $i(i=1,2, \cdots, n)$, we have $a_{i} \leqslant \alpha \leqslant \beta \leqslant b_{i}$. Each $\Delta_{i}$ contains $\alpha$, so there are no non-intersecting interval pairs $\left(\Delta_{i}, \Delta_{i+1}\right)$.

If $\beta<\alpha$, then there exists $i \in\{1,2, \cdots, n\}$ such that $\beta=b_{i}$, and $a_{i}<b_{i}=\beta<\alpha=a_{2}<b_{2}$. Thus, $\Delta_{2}$ and $\Delta_{i}$ are non-intersecting. Since $\Delta_{2}$ intersects with all other intervals except $\Delta_{1}$ and $\Delta_{3}$, $\Delta_{2}$ and $\Delta_{i}$ are non-intersecting if and only if $i=1$ or $i=3$. By symmetry, assume $i=3$, then $\beta=b_{3}$. Since each of $\Delta_{4}, \Delta_{5}, \cdots, \Delta_{n}$ intersects with $\Delta_{2}$, for $i=4,5, \cdots, n$, we have $a_{i} \leqslant \alpha \leqslant b_{i}$. Therefore, $\alpha \in \Delta_{4} \cap \Delta_{5} \cap \cdots \cap \Delta_{n}$, i.e., $\Delta_{4} \cap \Delta_{5} \cap \cdots \cap \Delta_{n} \neq \varnothing$. Similarly, $\Delta_{5}, \cdots, \Delta_{n}, \Delta_{1}$ all intersect with $\Delta_{3}$. Thus, $\beta \in \Delta_{5} \cap \cdots \cap \Delta_{n} \cap \Delta_{1}$, i.e., $\Delta_{5} \cap \cdots \cap \Delta_{n} \cap \Delta_{1} \neq \varnothing$. Therefore, the remaining interval pairs $(\Delta_{1}, \Delta_{2})$, $(\Delta_{2}, \Delta_{3})$, $(\Delta_{3}, \Delta_{4})$ are the possible non-intersecting interval pairs. Thus, there are at most 3.

By the proposition, the maximum value of the positive integer $n$ that satisfies the condition is 6.