I claim that the answer is $2$. For each region $i$, assign a number $a_i$ such that all points in region $i$ is contained inside exactly $a_i$ of the $2019$ circles. I claim that no region $j$ adjacent to $i$ can have $a_i\equiv a_j\pmod{2}$. For the sake of contradiction, suppose this is not the case. Then for some adjacent regions $i$ and $j$, $a_i$ and $a_j$ have the same parity. But they are separated by only $1$ arc! That means this $1$ arc must have been shared by an even number of distinct circles, which is clearly impossible (each arc can only belong to one distinct circle). Thus the claim has been proven. Now, color all regions $i$ with even $a_i$ red, and color all regions $j$ with odd $a_j$ blue. From the claim above, this coloring is satisfactory, and obviously $1$ color is not enough for the map, so we are done.