The answer is $2020\binom{4038}{2019}$.
We show that for all sequences with $n$ zeros and $n$ ones, the sum of all continuous runs is given by $(n+1)\binom{2n}{n}$.

We first count the number of changes, which means a pair of consecutive digits which is $01$ or $10$. There are $2n-1$ positions for which changes may occur. For each fixed position, there are $2$ ways to choose the pair of digits for this change, and $\binom{2n-2}{n-1}$ ways to put the remaining $n-1$ zeros and $n-1$ ones. Hence, there are
$$2(2n-1)\binom{2n-2}{n-1}$$
changes in total.

Now, for each sequence, the number of runs is $1$ more than the number of changes. Since there are $\binom{2n}{n}$ sequences in total, the total number of runs is
$$\begin{aligned} & 2(2n-1)\binom{2n-2}{n-1} + \binom{2n}{n} \\ &= 2n\binom{2n-1}{n} + \binom{2n}{n} = 2n\binom{2n-1}{n-1} + \binom{2n}{n} \\ &= n\binom{2n}{n} + \binom{2n}{n} = (n+1)\binom{2n}{n}, \end{aligned}$$
using the fact $k\binom{m}{k} = m\binom{m-1}{k-1}$.