What is the sum of the digits of the integer equal to $3 \times 10^{500}-2022 \times 10^{497}-2022$ ?
We can rewrite the given expression as follows:

$$
\begin{aligned}
3 \times 10^{500}-2022 \times 10^{497}-2022 & =3000 \times 10^{497}-2022 \times 10^{497}-2022 \\
& =(3000-2022) \times 10^{497}-2022 \\
& =978 \times 10^{497}-2022 \\
& =\left(978 \times 10^{497}-1\right)-2021
\end{aligned}
$$

The integer $978 \times 10^{497}-1$ is the 500 -digit integer with leading digits 977 followed by 4979 's. Thus, the digits of the given integer, from left to right, are 9, 7 , and 7 followed by $497-4=493$ 9 's, and the final four digits are $9-2=7,9-0=9,9-2=7$, and $9-1=8$.

Therefore, the sum of the digits of the integer is

$$
9+7+7+(493 \times 9)+7+9+7+8=4491
$$

ANSWER: 4491