One more of those

Let $S_n=\sum_{k=0}^{\infty}{2020 \choose kn}$ . Since ${2020 \choose j} = {2020 \choose 2020-j}$ , we have $S_8=\sum_{k=0}^{\infty} {2020 \choose 8k+4}$ , so that $S_8=\frac{S_4}{2}.$

Now we use a roots of unity filter , $S_4=\frac{1}{4}\left((1+1)^{2020}+(1+i)^{2020}+(1-1)^{2020}+(1-i)^{2020}\right)$ $=\frac{1}{4}\left(2^{2020}+(2i)^{1010}+(-2i)^{1010}\right)=\frac{1}{4}(2^{2020}-2^{1011})=2^{2018}-2^{1009}$

Finally $S=S_8-1=\frac{S_4}{2}-1=2^{2017}-2^{1008}-1$ . Now $2^{2017}-1$ is all ones to base 2, and $S$ has just $\boxed{1}$ zero.

Let $f(x) = (1+x)^{2020}$ , then we have:

$S = \displaystyle \sum_{k=1}^{2020} \begin{pmatrix} 2020 \\ 8k \end{pmatrix} \\ \quad = \displaystyle \sum_{k=0}^{2020} \begin{pmatrix} 2020 \\ 8k \end{pmatrix} - 1 \quad \quad \small \color{#3D99F6}{\text{Let } \omega^8 = 1 \text{, the eighth root of unity.}} \\ \quad = \frac{1}{8} \left(f(1) + f(1+\omega) + f(1+\omega^2) + f(1+\omega^3) + f(1+\omega^4) \\ \quad \quad + f(1+\omega^5) + f(1+\omega^6) + f(1+\omega^7) \right) - 1 \\ \quad = \frac{1}{8} \left((1+1)^{2020} + \left(1+\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right)^{2020} + (1+i)^{2020} + \left(1-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right)^{2020} + (1-1)^{2020} \\ \quad \quad + \left(1-\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\right)^{2020} + (1-i)^{2020} + \left(1+\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\right)^{2020} \right) - 1 \\ \quad = \frac{1}{8} \left(2^{2020} + 2^{505}(1+ \sqrt{2})^{2020}i - 2^{1010} - 2^{505}(1- \sqrt{2})^{2020}i + 0 \\ \quad \quad + 2^{505}(1- \sqrt{2})^{2020}i -2^{1010} - 2^{505}(1+ \sqrt{2})^{2020}i \right) - 1 \\ \quad = \frac{1}{8} \left(2^{2020} - 2^{1011} \right) - 1 \\ \quad = 2^{2017} - 2^{1008} - 1 \\ \quad = (1\underbrace{000...000}_{2017 \text{ zeroes}})_2 - (1\underbrace{000...000}_{1008 \text{ zeroes}})_2 - (1)_2 \\ \quad = (\underbrace{111...111}_{2017 \text{ ones}} \underbrace{000...000}_{1008 \text{ zeroes}})_2 - (1)_2 \\ \quad = (\underbrace{111...111}_{2016 \text{ ones}} \boxed{0} \underbrace{111...111}_{1008 \text{ ones}})_2$

$\Rightarrow$ There is only $\boxed{1}$ zero in $S$ in base-2.