I Did See You Before!

We can directly check $a_n = 10000 + 103(n-1)$

Then $a_m \equiv a_n \pmod{2017} \\ 10000 + 103(m-1) \equiv 10000 + 103(n-1) \pmod{2017} \\ 103m \equiv 103n \pmod{2017}$

Since $2017$ is prime and $103\not\equiv 0 \pmod{2017}$ (or more generally, since $\gcd(103,2017) = 1$ ), we can cancel it from both sides to give $m \equiv n \pmod{2017},$ so there exists some integer $k$ such that $m = n + 2017k$

Since $m > n$ , $k$ is positive, so $m$ is minimized when $n=k=1$ , giving $m = 1 + 2017 = \boxed{2018}$