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We need to find the sum of all the numbers on the first $2019$ rows of Pascal's Triangle that are divisible by $5$.

The $n$th row of Pascal's Triangle has $n$ numbers, and the $k$th number in the $n$th row is $\binom{n-1}{k-1}$. Therefore, we need to find the sum of all the numbers of the form $\binom{n-1}{k-1}$ where $1\leq n\leq 2019$ and $1\leq k\leq n$ that are divisible by $5$.

Note that $\binom{n-1}{k-1}\equiv 0\pmod{5}$ if and only if both $n$ and $k$ are not divisible by $5$. Therefore, the sum we want is equal to $$\sum_{n\equiv 0,1,2,3,4\pmod{5}}\sum_{k=1}^n\binom{n-1}{k-1}.$$ Using the identity $\sum_{k=0}^n\binom{n}{k}=2^n$, we get that the inner sum is equal to $\binom{n-1}{0}+\binom{n-1}{1}+\cdots+\binom{n-1}{n-1}=2^{n-1}$. Thus, our desired sum is equal to $$\sum_{n\equiv 0,1,2,3,4\pmod{5}}2^{n-1}\cdot n.$$ We can evaluate this sum as follows: \begin{align*} \sum_{n=1}^{2019}2^{n-1}\cdot n&=\sum_{n=0}^{2018}2^{n}\cdot (n+1)\ &=\sum_{n=0}^{2018}(n+1)\cdot 2^{n+1}-\sum_{n=0}^{2018}2^{n+1}\ &=\sum_{n=0}^{2018}(n+1)\cdot 2^{n+1}-2^{2020}+2\ &=(2^1+2^2)+(2^2+2^3\cdot 2)+(2^3+2^4\cdot 3)+\cdots+(2^{2019}+2^{2020}\cdot 2019)-2^{2020}+2\ &=2^{2020}-2+2^2+2^3+2^4+\cdots+2^{2019}\cdot 2019\ &=2^{2020}-2+(2^{2020}-2^{2019})\ &=2^{2021}-2^{2019}-2. \end{align*}Therefore, the answer is $\boxed{2^{2021}-2^{2019}-2}$