Don't overthink

This problem basically asks for the area bounded by the graph of

$|x+y| + |x-y| = 2017$

which, graphically, is

a square centered at the origin with side length $2017$ . And thus, we can easily get its area, and that would be $2017^2 = \boxed{4068289}$

We can actually generalize the graph of a square centered at the origin for any side length $k$ . That would be $|x+y| + |x-y| = k$ And the "unit square" will have the equation with $k=1$ .

Since the obvious graphical solution was posted, here's a computational solution. Let's first start with the inequality $|x-y| + |x+y| \leq 2017.$ If $x>y$ then we have $x \leq \frac{2017}{2}$ . If $x<y$ such that $x+y>0$ , then we get $y \leq \frac{2017}{2}$ . If $x< y$ such that $x+y<0$ , then we get $x \geq \frac{-2017}{2}$ . If $y<x$ such that $x+y<0$ , then we get $y \geq \frac{-2017}{2}$ . Using this we can construct the integral $A = \int_{-\frac{2017}{2}}^{\frac{2017}{2}}\int_{-\frac{2017}{2}}^{\frac{2017}{2}} \,dx\,dy$ $A = 4\Big(\frac{2017}{2}\Big)^2 = 2017^2 = \boxed{4068289}.$