Problem 1516

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. $k+1=2 \Rightarrow x=1009=y$.