Let $m$ be the initial integer Mary has picked. Because
$$2017x + 2 \equiv 2 - x \pmod{2018} \text{ and}$$
$$x + 2019 \equiv x + 1 \pmod{2018},$$
the numbers produced (mod $2018$) are
$$m \xrightarrow{M} 2 - m \xrightarrow{P} 3 - m \xrightarrow{M} 2 - (3 - m) = m - 1 \xrightarrow{P} m$$
where $M \xrightarrow{\cdot} P$ indicates Mary's turn and $\xrightarrow{\cdot}$ indicates Pat's turn. We see that the values produced (mod $2018$) form a cycle of length $4$. Hence, Mary wins exactly when none of the numbers $m, 2-m, 3-m, m-1$ is divisible by $2018$, i.e., if $m$ does not have remainder $0$, $1$, $2$ or $3$ on division by $2018$. The smallest integer greater than $2017$ that satisfies this condition is $2018 + 4 = 2022$.