Let $A$ be the 221-digit number all of whose digits are 9s. What is the sum of the digits of $A^{2}$?
Let $A_{n}$ denote the $n$-digit number consisting entirely of 9s.

Then $A_{n}=10^{n}-1$ and thus

$$
A_{n}^{2}=10^{2 n}-2 \cdot 10^{n}+1=10^{n}\left(10^{n}-2\right)+1
$$

The last digit of $10^{n}-2$ is 8, and the remaining $(n-1)$ digits are 9. When this number is multiplied by $10^{n}$, the sum of the digits does not change, and since the resulting number ends in zero, adding 1 increases the sum of the digits by 1.

Therefore, the sum of the digits of $A_{n}^{2}$ is $(n-1) \cdot 9+8+1=9 n$. In our case, $n=221$, so the sum of the digits of $A^{2}$ is $9 \cdot 221=1989$.

Remarks. 1. Essentially, it also becomes clear from the above solution that

$$
A_{n}^{2}=\underbrace{99 \ldots 9}_{n-1} 8 \underbrace{00 \ldots 0}_{n-1} 1
$$

2. In the paper by Levente Koncz (Bp. Árpád Gymn. II.), it is shown that for $n \geqq k$, the sum of the digits of the product $A_{n} A_{k}$ is also $9 n$.