Example 5 Does there exist a natural number $n$, such that the sum of the digits of $n^{2}$ equals: (1) 2003; (2) 2002.
Solution: (1) Since the remainder of any natural number divided by 9 is equal to the remainder of the sum of its digits divided by 9, and $n^{2} \equiv$ $0,1,4,7(\bmod 9)$, while $2003 \equiv 5(\bmod 9)$, there does not exist such a natural number $n$ that the sum of the digits of $n^{2}$ equals 2003.
$$
\begin{array}{l}
\text { (2) Let } n=10^{22}-3 \text {, then } \\
n^{2}=10^{44}-6 \times 10^{22}+9 \\
=\underbrace{99 \cdots 94}_{21} \underbrace{00 \cdots 09}_{21} \text {. } \\
\end{array}
$$

Thus, the sum of the digits is $9 \times 222+4=2002$.
The first part of the question uses the method of elimination; the second part uses the method of providing an example as mentioned in Example 1.