, 2019

Solution:

We have
$$x^{2}-y^{2} = \max(x, y) - \min(x, y) = |x-y|$$
Now if $x \neq y$, we can divide by $x-y$ to obtain $x+y= \pm 1$. Thus $x=y$ or $x+y= \pm 1$.

If $x=y$, we see that $2019 > x^{2} + y^{2} = 2x^{2}$, so we see that $-31 \leq x \leq 31$. There are 63 ordered pairs in this case.

In the second case, note that $|x| \geq |y|$ since $x^{2} - y^{2} = |x-y| \geq 0$. Since $x+y= \pm 1$, we cannot have $xy > 0$, so either $x \geq 0, y \leq 0$, or $x \leq 0, y \geq 0$. In the first case, $x+y=1$; in the second case, $x+y=-1$. Thus, the solutions for $(x, y)$ are of the form $(k, 1-k)$ or $(-k, k-1)$ for some $k>0$. In either case, we must have $k^{2} + (k-1)^{2} < 2019$, which holds true for any $1 \leq k \leq 32$ but fails for $k=33$. There are a total of $32 \cdot 2 = 64$ solutions in this case.

In summary, there are a total of $63 + 64 = 127$ integer solutions to the equation $x^{2} + \min(x, y) = y^{2} + \max(x, y)$ with $x^{2} + y^{2} < 2019$.