Setting $x=1$, $i$, $-1$ and $-i$, where $i=\sqrt{-1}$, in the binomial expansion of $(1+x)^{2020}$ yields $$\begin{array}{rrl} S_1: &(1+1)^{2020} =& \binom{2020}{0} + \binom{2020}{1} + \binom{2020}{2} + \binom{2020}{3} + \binom{2020}{4} + \cdots \\ S_i: & (1+i)^{2020} =& \binom{2020}{0} + i\binom{2020}{1} - \binom{2020}{2} - i\binom{2020}{3} + \binom{2020}{4} + \cdots \\ S_{-1}: & (1-1)^{2020} =& \binom{2020}{0} - \binom{2020}{1} + \binom{2020}{2} - \binom{2020}{3} + \binom{2020}{4} + \cdots \\ S_{-i}: & (1-i)^{2020} =& \binom{2020}{0} - i\binom{2020}{1} - \binom{2020}{2} + i\binom{2020}{3} + \binom{2020}{4} + \cdots \end{array}$$

Adding these four relation together gives $$\begin{array}{c} 2^{2020} + (1+i)^{2020} + 0 + (1-i)^{2020}= 4\times\left(\binom{2020}{0} +\binom{2020}{4} + \cdots \right)\end{array}$$

Meanwhile, $S_1 - S_{i} + S_{-1} - S_{-i}$ gives $$\begin{array}{c} 2^{2020} - (1+i)^{2020} + 0 - (1-i)^{2020}= 4\times\left(\binom{2020}{2} +\binom{2020}{6} + \cdots \right)\end{array}$$

It follows that $$\begin{array}{rl} S=& \frac{1}{4}\times 2\times \left((1+i)^{2020} + (1-i)^{2020}\right) \\ =& \frac{1}{2} \times \left((\sqrt{2}e^{\frac{\pi}{4}i})^{2020} + (\sqrt{2}e^{-\frac{\pi}{4}i})^{2020}\right) \\ =& 2^{1009}\times\left(e^{i\pi} + e^{-i\pi}\right) \\ = & 2^{1009}\times (-2) \\ =&\boxed{-2^{1010}}\end{array}$$