10.45 Let $S(n)$ denote the sum of the digits of the natural number $n$.
(1) Does there exist a natural number $n$ such that $n + S(n) = 1980$?
(2) Prove that among any two consecutive natural numbers, one can be expressed in the form $n + S(n)$, where $n$ is some natural number.
(14th All-Soviet Union Mathematical Olympiad, 1980)
[Solution] (1) From $n+S(n)=1980$, we know that $n$ is a four-digit number,
thus
$$
S(n) \leqslant 36 \text{. }
$$

Therefore, $n$ is in the form of $\overline{19 x y}$.
From the given condition, we have $1900+10 x+y+(1+9+x+y)=1980$.

Thus,
$$
\begin{array}{c}
11 x+2 y=70 . \\
5 \leqslant x \leqslant 6 .
\end{array}
$$

Since $x$ is even, we have $x=6, y=2$.
Therefore,
$$
n=1962 \text{. }
$$

We can verify that
$$
1962+S(1962)=1980 \text{. }
$$
(2) Let
$$
S_{n}=S(n)+n \text{. }
$$

If the unit digit of $n$ is 9, then $S_{n+1}2$. Choose the largest $N$ such that $S_{N}<m$ and the unit digit of $N$ is not 9, then
$$
S_{N+1} \geqslant m \text{. }
$$

Thus, $S_{N+1}=m$ or $S_{N+1}=m+1$. Therefore, one of $m$ and $m+1$ can be expressed in the form $S_{n}=n+S(n)$.