C/1 Find all real numbers $x$ that satisfy the equation
$$
\frac{x-2020}{1}+\frac{x-2019}{2}+\cdots+\frac{x-2000}{21}=\frac{x-1}{2020}+\frac{x-2}{2019}+\cdots+\frac{x-21}{2000},
$$
and simplify your answer(s) as much as possible. Justify your solution.
Solution 1. 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 2. 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 $x=2021$.