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 $$\begin{array}{rl}& \left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}+2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}+2018\right)\\ & ={\left(x+\frac{1}{y}\right)}^2+{\left(y+\frac{1}{x}\right)}^2+2018\left(x+\frac{1}{y}\right)+2018\left(y+\frac{1}{x}\right)\\ & \ge \frac{1}{2}{\left(x+\frac{1}{y}+y+\frac{1}{x}\right)}^2+2018\left(x+\frac{1}{y}+y+\frac{1}{x}\right)\\ & =\frac{1}{2}{u}^2+2018u,\end{array}$$where $u=x+\frac{1}{y}+y+\frac{1}{x}.$

By AM-GM, $$u=x+\frac{1}{x}+y+\frac{1}{y}\ge 2+2=4.$$The function $\frac{1}{2}{u}^2+2018u$ is increasing for $u\ge 4,$ so $$\frac{1}{2}{u}^2+2018u\ge \frac{1}{2}\cdot 4^2+2018\cdot 4=8080.$$Equality occurs when $x=y=1,$ so the minimum value is $\boxed{8080}.$