Don't check Primality

Proposition : Sum of any fraction and its inverse can not be an integer.

Consider a fraction $\dfrac{p}{q}$ .Now assume that $\dfrac{p}{q} + \dfrac{q}{p}=m$ where $m \in \mathbb{Z}$

So $p^2+q^2=mpq \\ \implies p^2-mpq+q^2=0$ .As $p$ is an integer so its discriminant should be an integer.

$D=\sqrt{m^2-4} \\ \implies m^2-D^2=4 \\ \implies (m+D)(m-D)=4$ .

From this the only solution is $D=0,m=2,$ .Putting $m=2$ yields $p=q$ which implies no fraction exists.

Now coming to the main problem.As stated that solution can be rational numbers we can assume

$x=\dfrac{a}{b} \,\text{and} \, y=\dfrac{c}{d}$ where $a,b,c,d \in \mathbb{Z};gcd(a,b)=gcd(c,d)=1;b,d \neq 0$

The expression becomes $\dfrac{a}{b} + \dfrac{c}{d} + \dfrac{b}{a}+ \dfrac{d}{c}=2019$

Case(1): if $x,y \in Z$ then inverses can not become an integer untill they are $1/2$ .But they cannt be half because it is not a solution.

Case(2):if $x=n\in Z; y=\frac{p}{q} \in Q$ and vice-versa..

$n+1/n+p/q+q/p=2019 \\ 1/n+p/q+q/p=2019-n=I$ .This will not give any solutions.

Case(3) When both $x,y \in Q$ then $\dfrac{a}{b} + \dfrac{c}{d} + \dfrac{b}{a}+ \dfrac{d}{c}=2019$

The sum of fraction and its inverse will not make any integer.So now consider $\dfrac{a}{b} + \dfrac{c}{d}=k \in Z$ and $\dfrac{b}{a}+ \dfrac{d}{c}=l \in Z$ .Multiply this:

$(\dfrac{a}{b} + \dfrac{c}{d})(\dfrac{b}{a}+ \dfrac{d}{c})=kl$

$1+ \dfrac{ad}{bc} +1 + \dfrac{bc}{ad}=kl \\ \dfrac{ad}{bc} + \dfrac{bc}{ad}=kl-2$ .But we proved earlier that fraction and sum with its inverse will never yields an integer.(Here fraction is $\dfrac{ad}{bc}$ )

if here $c|a$ and $b|d$ then $\dfrac{ad}{bc}$ will be an integer but $\dfrac{bc}{ad}$ will again become an fraction.

So there is no rationals $x,y$ exists.