Problem 835

Prove that for some integer $m(m>1)$, assuming there exists a positive integer $n$ such that equation (1) holds. Let
$$n=10^{k} a_{k}+10^{k-1} a_{k-1}+\cdots+10 a_{1}+a_{0} \text {, }$$

where $a_{k}, a_{k-1}, \cdots, a_{0}$ are non-negative integers no greater than 9, and $a_{k} \geqslant 1$.
Thus, $S(n)=a_{k}+a_{k-1}+\cdots+a_{0}$.
(1) $k<10^{m}-10$, contradiction.
(3) $k>m$.

It is easy to see that $n$ has at least one more digit than $10^{m}$.
$$\begin{array}{l} \text { By } a_{k}+a_{k-1}+\cdots+a_{m}+a_{m-1}+\cdots+a_{0} \\ =10^{k} a_{k}+10^{k-1} a_{k-1}+\cdots+10^{m}\left(a_{m}-1\right)+ \\ 10^{m-1} a_{m-1}+\cdots+a_{0}+10 \text {, } \end{array}$$

then $a_{k}=1, a_{k-1}=\cdots=a_{m}=a_{m-1}=\cdots=a_{1}=0$.
Thus, $n-S(n)=10^{k}-1>10^{m}-10$, contradiction.
In summary, the generalization holds.