(1) Does there exist a natural number $n$, such that $n+S(n)=1980$?
(2) Prove that among any two consecutive natural numbers, at least one can be expressed in the form $n+S(n)$ (where $n$ is another natural number).
(1) If there exists $n$ such that $n+S(n)=1980$, then by $S(n) \geqslant 1$ we know $n \leqslant 1979$. Since $n \leqslant 1979 < 1999$, it follows that $S(n) \leqslant 28$, hence $n \geqslant 1952$. By verifying the numbers between 1952 and 1979, we find that $1962+S(1962)=1980$.

(2) If the unit digit of $n$ is not 9, then $(n+1)+S(n+1)$ is 2 more than $n+S(n)$. If the unit digit of $n$ is 9, then $(n+1)+S(n+1)$ is less than $n+S(n)$.

Clearly, $n+S(n)$ is unbounded. For any natural number $m \geqslant 3$, when $n=1$, $n+S(n)<m$. As $n$ increases, there must be a point where $n+S(n) \geqslant m$. At the first occurrence of $n+S(n) \geqslant m$, since $n+S(n)$ is 2 more than $n-1+S(n-1)(<m)$, it follows that $n+S(n)=m$ or $n+S(n)=m+1$ must hold. That is, at least one of any two consecutive natural numbers $m$ and $m+1$ can be expressed in the form $n+S(n)$ (for $m=1,2$, this conclusion is obviously true).