Solution:

We prove the contrapositive and assume for this purpose that every rectangle of the decomposition has at least one point in common with the boundary of the square. By the hypotheses, it then has at least one of its sides in common with the boundary of the square. Hence the boundary of the square can be assigned piecewise, in a unique way, to the rectangles of the decomposition. Two non-connected pieces cannot belong to the same rectangle; otherwise they would have to lie opposite one another, and on the other sides of this rectangle lines would intersect the interior of the square without running through the interior of any rectangle.

The number $n$ of these pieces therefore coincides with the number $n$ of rectangles. At each of the $n$ endpoints of the pieces, on the boundary of the square, two rectangles meet and each has a corner there. Four further corners of the rectangles coincide with the corners of the square, so that altogether $2 n+4$ rectangle corners lie on the boundary of the square. Thus $4 n-(2 n+4)=2 n-4$ corners remain for the interior of the square. Now we consider one of the $n$ points on the boundary of the square at which two rectangles meet. The line perpendicular to the respective side of the square through this point $P$

runs, within the interior of the square, at first along one side each of the two rectangles. In order for this line to also run through the interior of a rectangle within the square, it must intersect a rectangle side that runs parallel to the side of the square from which we started. At this branching point $Q$, the two rectangles whose boundary the line previously formed (these need no longer be the same two rectangles as those with corner point $P$) each have a corner point. Such a point $Q$ exists for every starting point $P$, and two different starting points cannot have the same branching point $Q$. Hence at least $2 n$ rectangle corners must lie in the interior of the square, contradicting the maximum number $2 n-4$ determined above. Therefore there always exists a rectangle that has no point in common with the boundary of the square.