1. Does there exist a perfect square whose digit sum is 1993?
[Solution] It exists. Because
$$\underbrace{99 \cdots 99}_{n \uparrow 9} 7^{2}=\underbrace{99 \cdots 99}_{n \uparrow 9} 4 \underbrace{00 \cdots 00}_{n \uparrow 0} 9,$$

and when $n=220$, the perfect square
$$\underbrace{99 \cdots 99}_{220 \uparrow 9} 4 \underbrace{00 \cdots 0}_{220 \uparrow 0} 9$$

has a digit sum of 1993.
Note: Another set of perfect squares can also be used:
$$\begin{array}{l}
33 \cdots 334^{2}=11 \cdots 1155 \cdots 556 . \\
{ }_{n \uparrow 3}{ }^{n+1 \uparrow 1} \quad n \text { s }
\end{array}$$

When $n=331$, the perfect square
$$\begin{array}{cc}
11 \cdots 11 & 55 \cdots 556 \\
332 \uparrow 1 & 331 \uparrow 5
\end{array}$$

has a digit sum of 1993.