Align the bottom left corner of the big rectangle with a checkerboard whose squares have side 1/2. Let the bottom left corner be black.
Two intuitive lemmas:
If both sides of a rectangle aren't whole, then the black area covered will be greater than white area covered.
On the contrary, if the black and white areas are equal, then the rectangle must have an integer side.
Every small rectangle has at least a integer side, so it covers equal areas of black and white.
So, the big rectangle (which is the sum of all small rectangles) must cover equal areas of black and white, thus must have an integer side.
Credits: I've read this evidence somewhere few years ago, I don't remember the author though. After googling a bit, I've found out that Stan Wagon has posted 14 solutions of this problem, incredible!