The answer is 898.

Indeed, let $\omega = e^{\frac{2\pi i}{3}}$ be a cube root of unity. By the binomial theorem, we have
$$(1 + 1)^n = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n},$$
$$(1 + \omega)^n = \binom{n}{0} + \binom{n}{1}\omega + \dots + \binom{n}{n}\omega^n,$$
$$(1 + \omega^2)^n = \binom{n}{0} + \binom{n}{1}\omega^2 + \dots + \binom{n}{n}\omega^{2n}.$$
It is known that $1 + \omega^k + \omega^{2k} = 0$ for any $3 \nmid k$, and $1 + \omega^k + \omega^{2k} = 3$ for any $3 \mid k$.

Thus, adding up the above 3 equations, we obtain
$$\begin{aligned} & 3\left(\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \dots + \binom{n}{3\left\lfloor \frac{n}{3} \right\rfloor}\right) \\ &= (1+1)^n + (1+\omega)^n + (1+\omega^2)^n \\ &= 2^n + (-\omega^2)^n + (-\omega)^n \\ &= \begin{cases} 2^n + 2 & \text{if } n \equiv 0 \pmod{6}, \\ 2^n + 1 & \text{if } n \equiv 1,5 \pmod{6}, \\ 2^n - 1 & \text{if } n \equiv 2,4 \pmod{6}, \\ 2^n - 2 & \text{if } n \equiv 3 \pmod{6}. \end{cases} \end{aligned}$$
Therefore, we have $f(n) = \frac{1}{3}$ if $3 \nmid n$, and $f(n) = \frac{2}{3}$ if $3 \mid n$. As $\lfloor \frac{2021}{3} \rfloor = 673$, we have
$$f(1) + f(2) + \dots + f(2021) = 673\left(\frac{2}{3}\right) + (2021 - 673)\left(\frac{1}{3}\right) = 898.$$