58th Ukrainian National Mathematical Olympiad

Since $(x,y)\cdot [x,y]=xy$, then the first equation of the system can be re-written as $$\frac{xy}{(x,y)}+(x,y)=2018,$$ which gives us quadratic equation with respect to $(x,y)$: $$(x,y{)}^2-2018(x,y)+xy=0.$$ Its discriminant, also taking into account the second equation of the system, is $$D=2018^2-4xy=2018^2-4x(2018-x)=(2x-2018{)}^2.$$ Which gives $x=\frac{2018\pm (2x-2018)}{2}\Rightarrow x_1=x$ and $x_2=2018-x=y$.

Hence, one of the numbers equals GCD, which means it divides the other number. E.g., let $x\le y$, i.e. $y=kx$, then $(k+1)x=2018$. Hence, these are the possible cases: Case 1. $k+1=2018\Rightarrow x=1$ and $y=2017$. Case 2. $k+1=1009\Rightarrow x=2$ and $y=2016$. Case 3. $k+1=2\Rightarrow x=1009=y$.