By the geometric series formula$$1 + 2019 + 2019^2 + 2019^3 +.. + 2019^{n-1}=\frac{2019^n-1}{2018}.$$If $n$ is even, by LTE, we have$$v_2(\frac{2019^n-1}{2018})=v_2(2019-1)+v_2(2019+1)+v_2(n)-1-v_2(2018)=v_2(n)+1.$$So the maximum power of $2$ if $n$ is even is simply $v_2(n)+1.$ If $n$ is odd then the whole expression is $1 \pmod{2}.$ So the answer is $\boxed{0}$ if $n$ is odd, and the answer is $\boxed{v_2(n)+1}$ if $n$ is even. $\blacksquare$