49th International Mathematical Olympiad Spain

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 $B_1,\dots ,B_n$ 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 $1\le k\le n$.

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 $k=1,\dots ,n$. Hence $(I_k,I_{k+1})$ or $(J_k,J_{k+1})$ is a pair of disjoint intervals, $1\le k\le n$. So there are at least $n$ pairs of disjoint intervals among $\left(I_1,I_2\right),\dots ,\left(I_{n-1},I_n\right),\left(I_n,I_1\right);\left(J_1,J_2\right),\dots ,\left(J_{n-1},J_n\right),\left(J_n,J_1\right)$.

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 $\left(I_1,I_2\right),\dots ,\left(I_{n-1},I_n\right),\left(I_n,I_1\right)$ are disjoint, and the same holds for $\left(J_1,J_2\right),\dots ,\left(J_{n-1},J_n\right),\left(J_n,J_1\right)$. Consequently $n\le 3+3=6$, as stated. Thus we are left with the claim and its justification.

Claim. Let ${\Delta }_1,{\Delta }_2,\dots ,{\Delta }_n$ 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 $k=1,\dots ,n$.

Proof. Denote ${\Delta }_k=[a_k,b_k],1\le k\le n$. Let $\alpha =\max \left(a_1,\dots ,a_n\right)$ be the rightmost among the left endpoints of ${\Delta }_1,\dots ,{\Delta }_n$, and let $\beta =\min \left(b_1,\dots ,b_n\right)$ be the leftmost among their right endpoints. Assume that $\alpha =a_2$ without loss of generality.

If $\alpha \le \beta$ then $a_i\le \alpha \le \beta \le b_i$ 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 ${\Delta }_4,\dots ,{\Delta }_n$ intersects ${\Delta }_2$, we have $a_i\le \alpha \le b_i$ for $i=4,\dots ,n$. Therefore $\alpha \in {\Delta }_4\cap \dots \cap {\Delta }_n$, in particular ${\Delta }_4\cap \dots \cap {\Delta }_n\ne \varnothing$. Similarly, ${\Delta }_5,\dots ,{\Delta }_n,{\Delta }_1$ all intersect ${\Delta }_3$, so that ${\Delta }_5\cap \dots \cap {\Delta }_n\cap {\Delta }_1\ne \varnothing$ as $\beta \in {\Delta }_5\cap \dots \cap {\Delta }_n\cap {\Delta }_1$. 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.