The equation can be rewritten as $n=\frac{4xy}{x+y}$. Let $x=da,y=db$ where $d=\gcd (x,y)$ such that $(a,b)=1$. Then $n=\frac{4dab}{a+b}$. Notice that $(a,a+b)=(b,a+b)=1$, and thus $ab|n$. Note that $a,b$ can't both be even, so WLOG let $a$ be odd. If $a\equiv 3\text{(mod 4)}$ then we're done, so let $a\equiv 1\text{(mod 4)}$. But since $n$ is odd, we must have $4|a+b\implies b\equiv 3\text{(mod 4)}$, so we're done.