jmc

Let $BC=x$ and $CD=AD=y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that$$x^2+(y-2{)}^2=y^2.$$Simplifying yields $x^2-4y+4=0$, so $x^2=4(y-1)$. Thus, $y$ is one more than a perfect square. The perimeter $p=2+x+2y=2y+2\sqrt{y-1}+2$ must be less than 2015. Simple calculations demonstrate that $y=31^2+1=962$ is valid, but $y=32^2+1=1025$ is not. On the lower side, $y=1$ does not work (because $x>0$), but $y=1^2+1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y=n^2+1$ for $1\le n\le 31$), and so our answer is $\boxed{31}$