Problem 1983

The maximum number of such boxes is $6$. One example is shown in the figure.

Now we show that $6$ is the maximum. Suppose that boxes $B_{1}, \ldots, B_{n}$ satisfy the condition. Let the closed intervals $I_{k}$ and $J_{k}$ be the projections of $B_{k}$ onto the $x$- and $y$-axis, for $1 \leq k \leq n$.

If $B_{i}$ and $B_{j}$ intersect, with a common point $(x, y)$, then $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$. So the intersections $I_{i} \cap I_{j}$ and $J_{i} \cap J_{j}$ are nonempty. Conversely, if $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$ for some real numbers $x, y$, then $(x, y)$ is a common point of $B_{i}$ and $B_{j}$. Putting it around, $B_{i}$ and $B_{j}$ are disjoint if and only if their projections on at least one coordinate axis are disjoint.

For brevity we call two boxes or intervals adjacent if their indices differ by $1$ modulo $n$, and nonadjacent otherwise.

The adjacent boxes $B_{k}$ and $B_{k+1}$ do not intersect for each $k=1, \ldots, n$. Hence $(I_{k}, I_{k+1})$ or $(J_{k}, J_{k+1})$ is a pair of disjoint intervals, $1 \leq k \leq n$. So there are at least $n$ pairs of disjoint intervals among $\left(I_{1}, I_{2}\right), \ldots,\left(I_{n-1}, I_{n}\right),\left(I_{n}, I_{1}\right);\left(J_{1}, J_{2}\right), \ldots,\left(J_{n-1}, J_{n}\right),\left(J_{n}, J_{1}\right)$.

Next, every two nonadjacent boxes intersect, hence their projections on both axes intersect, too. Then the claim below shows that at most $3$ pairs among $\left(I_{1}, I_{2}\right), \ldots,\left(I_{n-1}, I_{n}\right),\left(I_{n}, I_{1}\right)$ are disjoint, and the same holds for $\left(J_{1}, J_{2}\right), \ldots,\left(J_{n-1}, J_{n}\right),\left(J_{n}, J_{1}\right)$. Consequently $n \leq 3+3=6$, as stated. Thus we are left with the claim and its justification.

Claim. Let $\Delta_{1}, \Delta_{2}, \ldots, \Delta_{n}$ be intervals on a straight line such that every two nonadjacent intervals intersect. Then $\Delta_{k}$ and $\Delta_{k+1}$ are disjoint for at most three values of $k=1, \ldots, n$.

Proof. Denote $\Delta_{k}=[a_{k}, b_{k}], 1 \leq k \leq n$. Let $\alpha=\max \left(a_{1}, \ldots, a_{n}\right)$ be the rightmost among the left endpoints of $\Delta_{1}, \ldots, \Delta_{n}$, and let $\beta=\min \left(b_{1}, \ldots, b_{n}\right)$ be the leftmost among their right endpoints. Assume that $\alpha=a_{2}$ without loss of generality.

If $\alpha \leq \beta$ then $a_{i} \leq \alpha \leq \beta \leq b_{i}$ for all $i$. Every $\Delta_{i}$ contains $\alpha$, and thus no disjoint pair $(\Delta_{i}, \Delta_{i+1})$ exists.

If $\beta<\alpha$ then $\beta=b_{i}$ for some $i$ such that $a_{i}<b_{i}=\beta<\alpha=a_{2}<b_{2}$, hence $\Delta_{2}$ and $\Delta_{i}$ are disjoint. Now $\Delta_{2}$ intersects all remaining intervals except possibly $\Delta_{1}$ and $\Delta_{3}$, so $\Delta_{2}$ and $\Delta_{i}$ can be disjoint only if $i=1$ or $i=3$. Suppose by symmetry that $i=3$; then $\beta=b_{3}$. Since each of the intervals $\Delta_{4}, \ldots, \Delta_{n}$ intersects $\Delta_{2}$, we have $a_{i} \leq \alpha \leq b_{i}$ for $i=4, \ldots, n$. Therefore $\alpha \in \Delta_{4} \cap \ldots \cap \Delta_{n}$, in particular $\Delta_{4} \cap \ldots \cap \Delta_{n} \neq \emptyset$. Similarly, $\Delta_{5}, \ldots, \Delta_{n}, \Delta_{1}$ all intersect $\Delta_{3}$, so that $\Delta_{5} \cap \ldots \cap \Delta_{n} \cap \Delta_{1} \neq \emptyset$ as $\beta \in \Delta_{5} \cap \ldots \cap \Delta_{n} \cap \Delta_{1}$. This leaves $(\Delta_{1}, \Delta_{2})$, $(\Delta_{2}, \Delta_{3})$ and $\left(\Delta_{3}, \Delta_{4}\right)$ as the only candidates for disjoint interval pairs, as desired.