i) How do we find all $x,y ∈ \mathbb Q^+$ such that $x+y+ \dfrac1{x} +\dfrac1{y}$ is an integer ?
ii)Let $p$ be an odd prime and $q=\dfrac{p-1}2$ , then how do we find all $x,y ∈ \mathbb Q^+$ such that $x^q+y^q+ \dfrac1{x^q} +\dfrac1{y^q}$ is an integer ?
( I only know that if $x,y ∈ \mathbb Q^+$ , then $x+y+ \dfrac1{x} +\dfrac1{y}+2$ is not divisible by $4$ and if $p$ is an odd prime then $x^q+y^q+ \dfrac1{x^q} +\dfrac1{y^q}$ is not divisible by $p$)
It may help to consider $x = \frac{a}{b}$, $y = \frac{c}{d}$ with $a,b$ and $c,d$ pairs of coprime positive numbers. Then, your condition is equivalent to $$abcd \mid (a^2+b^2)cd+(c^2+d^2)ab.$$ Since $a,b$ are coprime, we find $ab \mid cd$ and same reasoning leads to $cd \mid ab$, thus $ab=cd$. So an elementary-number-theory equivalent of your condition would be: $$ab=cd \\ ab \mid a^2+b^2+c^2+d^2.$$ I am not sure if there is any closed form for the set of solutions. Example solution are $(a,b,c,d) = (1,2,1,2)$ and $(a,b,c,d)=(2,4981,17,568)$.
You get similar equations for the generalized case.