1. Is there a natural number $n$ such that the sum of the digits of $n^{2}$ equals 1983? Equals 1984?
Solution: The remainder of $n$ divided by 9 may be $0,1,2,3$, $4,5,6,7,8$, so the remainder of $n^{2}$ divided by 9 may be $0,1,4,7$. Therefore, the remainder of the sum of the digits of $n^{2}$ divided by 9 can only be $0,1,4,7$.

Since 1983 divided by 9 leaves a remainder of 3, the product of the digits of $n^{2}$ cannot be 1983.
The sum of the digits of $a^{2}$ can be 1984. In fact, when
$$
n=10^{22}-3
$$

then,
$$
\begin{aligned}
n^{2} & =10^{44}-6 \times 10^{21}+9 \\
& 219 \text{ nines} \\
& =99 \cdots 940 \cdots 09,
\end{aligned}
$$

the sum of its digits is exactly 1984.