TITLE: Venn diagram with rectangles for $n > 3$
QUESTION [11 upvotes]: Is it possible to draw a Venn diagram with rectangles for $n > 3$? If yes, is it possible for all $n\in\mathbb{N}$? If no, up to which $n$ is it possible?
Furthermore, the rectangles should have the following properties

same width and height
axis-aligned (not rotated)

REPLY [6 votes]: Suppose you have a Venn diagram with $n$ axis-aligned rectangles of the same width and height. By a scaling of the axes, we may assume the rectangles are all squares. Let's show that it's not possible to have a Venn diagram with $n$ axis-aligned squares for $n\ge4$, even if the squares are allowed to be of different sizes.
Suppose there were such a Venn diagram. By shrinking squares as necessary, we can assume the diagram is simple -- that is, each point of intersection of the perimeters of the squares is an intersection of just two squares. In that case, each pair of squares intersects exactly twice, for a total of $2{n\choose2}=n(n-1)$ vertices for the diagram, which implies a total of $2n(n-1)$ (topological) edges. From Euler's formula, $V-E+F=2$, we have
$$F=2+n(n-1)$$
and it's elementary to show that $2^n\gt2+n(n-1)$ for $n\ge4$.
Remark: Because each region in a Venn diagram is an intersection of interiors, the shrinking step(s) can clearly be engineered to leave an open set for each of the $2^n$ intersections, but it's conceivable you could wind up with duplicates of some of the intersections --i.e., the diagram may no longer be a Venn diagram.  However, that doesn't matter for the argument here, since it simply says $F\ge2^n$.
Additional remark: The scaling was done for convenience, since it's clear that overlapping squares cannot intersect at more than two points (if the intersections are simple). This is also true for rectangles with a common aspect ratio. It's not true, of course, in general:  a tall skinny rectangle can intersect a short fat one at $4$ points as easily as $2$.