jmc

By QM-AM, $$\sqrt{\frac{(x+\frac{1}{y}{)}^2+(y+\frac{1}{x}{)}^2}{2}}\ge \frac{(x+\frac{1}{y})+(y+\frac{1}{x})}{2},$$so $${\left(x+\frac{1}{y}\right)}^2+{\left(y+\frac{1}{x}\right)}^2\ge \frac{1}{2}{\left(x+\frac{1}{y}+y+\frac{1}{x}\right)}^2.$$Then $$\begin{array}{rl}& \left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}-2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}-2018\right)\\ & ={\left(x+\frac{1}{y}\right)}^2+{\left(y+\frac{1}{x}\right)}^2-2018\left(x+\frac{1}{y}\right)-2018\left(y+\frac{1}{x}\right)\\ & \ge \frac{1}{2}{\left(x+\frac{1}{y}+y+\frac{1}{x}\right)}^2-2018\left(x+\frac{1}{y}+y+\frac{1}{x}\right)\\ & =\frac{1}{2}{u}^2-2018u\\ & =\frac{1}{2}(u-2018{)}^2-2036162,\end{array}$$where $u=x+\frac{1}{y}+y+\frac{1}{x}.$

Equality occurs when $u=2018$ and $x=y.$ This means $x+\frac{1}{x}=1009,$ or $x^2-1009x+1=0.$ We can check that this quadratic has real roots that are positive, so equality is possible. Thus, the minimum value is $\boxed{-2036162}.$