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 $2019>x^{2}+y^{2}=2 x^{2}$, 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 $x y>0$, 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 $x^{2}+y^{2}<2019$.