We must have
$$f(n)=\lfloor \frac{n}{5}\rfloor+\lfloor \frac{n}{25}\rfloor+\lfloor \frac{n}{125}\rfloor+\ldots=2021$$But
if $n=8099$, then $f(8099)=1619+323+64+12+2=2020$.
if $n=8100$, then $f(8100)=1620+324+64+12+2=2022$.
And since $f(n)$ is non-decreasing on $n\in\mathbb N$, we see that there are $\boxed{\text{no}}$ such $n$.
$$\lfloor \frac{n}{5}\rfloor+\lfloor \frac{n}{25}\rfloor+\lfloor \frac{n}{125}\rfloor+ \cdots < \frac{n}{5} + \frac{n}{25} + \cdots$$$$ = \frac{\frac{n}{5}}{1 - \frac{1}{5}} = \frac{n}{4}$$$$\implies 2021 < \frac{n}{4}$$$$\implies n > 8084$$Now check numbers "near" 8084 to get an idea when there will be 2021 zeroes