It is a straightforward application of the root of unity filter that $f(n)$ is the absolute value is just $\frac{(1+\omega)^n+(1+\omega^2)^n}{3}=\frac{(-\omega^2)^n+(-\omega)^n}{3}$ where $\omega$ is a third root of unity. From here we see that $f(n)=\frac{2}{3}$ for $n$ a multiple of $3$ and $f(n)=\frac{1}{3}$ else. Hence the sum is $\frac{1}{3} \cdot \left(2021+\lfloor \frac{2021}{3}\rfloor\right)=898$.