The number $2016b + 1$ gives the same remainder upon division by $2017$ as $-b + 1$. Hence the remainders upon division by $2017$ are as in the sequence $b, -b+1, -(b+1)+1, \dots$. Since $-(b+1)+1 = b$, this sequence has period $2$, whence there are at most $2$ different remainders. Hence the number $a$ gives the win to Kati if and only if neither $a$ nor $-a+1$ is divisible by $2017$. Thus $a = 2017$ does not give a win, as it is divisible by $2017$, similarly for $a = 2018$, as $-2018+1$ is divisible by $2017$. But for $a = 2019$ none of $2019$ or $-2019+1$ is divisible by $2017$, giving a win.