jmc

If $n$ is even, then we can write $n=2m$ for some integer $m$. Substituting, $$\lfloor (2m{)}^2/4\rfloor -{\lfloor (2m)/2\rfloor }^2=m^2-m^2=0.$$Hence, $n$ must be odd; we can write $n=2m+1$ for some integer $m$. Substituting, $$\begin{array}{rl}& \lfloor (2m+1{)}^2/4\rfloor -{\lfloor (2m+1)/2\rfloor }^2\\ & =\lfloor (4m^2+4m+1)/4\rfloor -{\lfloor (2m+1)/2\rfloor }^2\\ & =\lfloor m^2+m+\frac{1}{4}\rfloor -{\lfloor m+\frac{1}{2}\rfloor }^2\\ & =m^2+m-m^2\\ & =m.\end{array}$$Thus, we find $m=2$ and $n=\boxed{5}$ as the unique integer solution.