Let $f(x)=(1+x)^{2020}$ and $\omega \ne 1$ be a complex root to the equation $x^3=1$. Then, applying the complex number method gives $$\begin{array}{rl} \displaystyle\sum_{n=1}^{673}a_n =& \displaystyle\frac{1}{3}\left(\omega^{2\times(3-0)}f(1)+\omega^{2\times(3-1)} f(\omega) + \omega^{2\times(3-2)}f(\omega^2)\right) \\ =&\displaystyle\frac{1}{3}\left(\omega\cdot 2^{2020} +\omega(1+\omega)^{2020} + \omega^2 (1+\omega^2)^{2020}\right) \end{array}$$

Setting $1+\omega=-\omega^2$, and $1+\omega^2=-\omega$ to the above relation will yield the final result as $$\boxed{\frac{2^{2020}+2}{3}}$$

The complex number method is introduced in the book  Power Calculation  and discussed in more detail in the practice by subject series  Binomial Expansion and Combinatorial Identities Practice .