This is basically the same as showing that $5^{2019}|4*5^{2018}*3+4*5^{2017}*3^{5}+..4*3^{5^{2018}}+3^{5^{2019}}$. Note that by LTE lemma, $v_5(3^{5^k}-3^{5^{k-1}})=k$. Thus $4*3^{5^{2018}}+3^{5^{2019}} \equiv 5*3^{5^{2018}}$ mod $5^{2019}$. Similarly, $5*3^{5^{2018}}+4*5*3^{5^{2017}} \equiv 25*3^{5^{2017}}$ mod $5^{2019}$. Going like this we see that this entire sum $4*5^{2018}*3+4*5^{2017}*3^{5}+..4*3^{5^{2018}}+3^{5*{2019}} \equiv 5^{2019}$ mod $5^{2019}$. Thus we get that $5^{2019}|4*5^{2018}*3+4*5^{2017}*3^{5}+..4*3^{5^{2018}}+3^{5*{2019}}$. Proved.
@below the first line is equal to the summation thing.
There are $4*5^{2018}$ terms coprime to $5^{2019}$, $4*5^{2017}$ terms such that their gcd with $5^{2019}$ is $5$, and so on till we reach $1$ term whose gcd with $5^{2019}$ is $5^{2019}$.