Using the condition given, we have $(m+n)^2 + (2018-m+n)^2 = 2018^2$. Let $m+n = u$, $2018-m+n = v$. It can be seen that $u,v > 0$. Since $u^2 + v^2 = 2018^2 \equiv 0 \pmod 4$, we get $u \equiv v \equiv 0 \pmod 2$. Let $u = 2u_1$, $v = 2v_1$. Then we have $u_1^2 + v_1^2 = 1009^2$. Since $(u_1, v_1, 1009)$ is a Pythagorean Triple, we get $1009 = d(r^2 + s^2)$, $(u_1, v_1) = (d(2rs), d(r^2-s^2))$ or $(u_1, v_1) = (d(r^2-s^2), d(2rs))$ for some positive integers $d, r, s$. Since $1009$ is a prime number, we obtain $d = 1$ and $r^2 + s^2 = 1009$. The last equation has solutions $(r,s) = (28,15), (15,28)$. Hence all solutions are $(u_1, v_1) = (840,559), (559,840)$ which correspond $(m,n) = (728,390), (1290,390).