Find the sum of the number of all continuous runs of all possible sequences with $2019$ ones and $2019$ zeros

Let us consider the general case with $n$ ones and $n$ zeroes.
We have to count the total number of runs $a_n$ in all the $\binom{2n}{n}$ sequences. For $i=1,\dots,2n-1$, the space between the $i$th digit and the $(i+1)$th digit marks the end of a run in $2\binom{2n-2}{n-1}$ cases (note that it does not depend on $i$). The space on the right side of the $2n$-th digit marks the end of the last run for all the $\binom{2n}{n}$ sequences. Since each run has a space on its right side, counting the runs is equivalent to count such spaces, that is
$$a_n=2\binom{2n-2}{n-1}\cdot (2n-1)+\binom{2n}{n}=(n+1)\binom{2n}{n}.$$ For $n=2019$ we find that $$a_{2019}=2020\binom{2\cdot 2019}{2019}=4040\binom{2\cdot 2019-1}{2018}=4040 \binom{4037}{2018}.$$