Numbers $S(n)$ and $S(n+1)$ have the following ratios: $S(n+1) = S(n)+1$, or $S(n) > S(n+1)$, and the last inequality can not be true for two numbers in a row. As $2018 = 2018 \cdot 1 = 1009 \cdot 2 = 1009 \cdot 1 \cdot 2$, and there are no other factor decompositions, that satisfy the condition, so the following cases are possible:
* $m=1$, $S(n)=2018$ and $S(n+1)=1$;
* $m=1$, $S(n)=1009$ and $S(n+1)=2$;
* $m=2$, $S(n)=1009$, $S(n+1)=1$ and $S(n+2)=2$.

If $S(n)=1009$ and $S(n+1)=1$, so we get contradiction as in the previous case. If $S(n)=1009$ and $S(n+1)=2$, so $n+1=100\dots0100\dots0$ or $n+1=200\dots0$. Then $n=100\dots0099\dots9$ and $n=199\dots9$. As $1009=112\cdot9+1$, so these variants are possible, and number $n$ has to have in the end 112 digits 9. The least value for $n$ is $n=\underbrace{199\dots9}_{112}$.