jmc

Call the height $h$ and the width $w$. We want to find the number of solutions to $2(w+h)=64$ or $w+h=32$. The solutions to this are $$\{(1,31),(2,30),\dots ,(16,16),\dots ,(31,1)\}.$$ There are 31 solutions to this, but we are double-counting all the rectangles for which $w\ne h$. There are 30 of these, so the total number of rectangles is $\frac{30}{2}+1=\boxed{16}$ rectangles.