Bay Area Mathematical Olympiad

Solution:
The number $x=2021$ works. Indeed, for $x=2021$, the left-hand side of the equation equals
$$\frac{2021-2020}{1}+\frac{2021-2019}{2}+\cdots+\frac{2021-2000}{21}=\frac{1}{1}+\frac{2}{2}+\cdots+\frac{21}{21}=\underbrace{1+1+\cdots+1}_{21}=21,$$
and the right-hand side of the equation equals the same number:
$$\frac{2021-1}{2020}+\frac{2021-2}{2019}+\cdots+\frac{2021-21}{2000}=\frac{2020}{2020}+\frac{2019}{2019}+\cdots+\frac{2000}{2000}=\underbrace{1+1+\cdots+1}_{21}=21.$$
Why is $x=2021$ the only solution? The equation is linear: after simplifying it, it can be put into the form $a x+b=0$. Such equations have a unique solution, namely, $x=-\frac{b}{a}$, as long as the coefficient $a$ of $x$ is not $0$. In our situation, $x$ is multiplied by
$$a=\underbrace{\left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{21}\right)}_{S_{1}}-\underbrace{\left(\frac{1}{2020}+\frac{1}{2019}+\cdots+\frac{1}{2000}\right)}_{S_{2}}.$$
However, each of the 21 fractions in the first sum $S_{1}$ is bigger than each of the 21 fractions in the second sum $S_{2}$. Thus, $S_{1}>S_{2}$ and $a>0$. Since $a \neq 0$, the equation has a unique solution, which we found earlier to be $x=2021$.

Solution:
If we subtract $21$ from both sides (i.e., we subtract $1$ from each of the $21$ fractions on both sides), we obtain
$$\frac{x-2021}{1}+\frac{x-2021}{2}+\cdots+\frac{x-2021}{21}=\frac{x-2021}{2020}+\frac{x-2021}{2019}+\cdots+\frac{x-2021}{2000}$$
or
$$\left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{21}\right)(x-2021)=\left(\frac{1}{2020}+\frac{1}{2019}+\cdots+\frac{1}{2000}\right)(x-2021)$$
which implies $x-2021=0$, or $x=2021$.