Just to add, that this comes from Baltic way 2002 contest. SEE http://www.math.su.se/bw2005/history.shtml

Let $x=\frac{a}b,y=\frac{c}d$ where $(a,b)$ and $(c,d)$ are pairs of coprime natural numbers. The equation becomes $\frac{a}b+\frac{c}d+\frac{b}a+\frac{d}c=3n$, or $(a^2+b^2)cd+(c^2+d^2)ab=3nabcd$. So $ab|cd$ and $cd|ab$, which implies $ab=cd$. Exactly 0 or 2 of $a,b,c,d$ are multiples of 3, so $a^2+b^2+c^2+d^2\equiv1\text{ or }2\pmod3$. But we have that $a^2+b^2+c^2+d^2=3nab$, which is impossible since $3\not|a^2+b^2+c^2+d^2$.