58th Ukrainian National Mathematical Olympiad

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 $n+1=100\dots 0100\dots 0$ or $n+1=200\dots 0$. Then $n=100\dots 0099\dots 9$ and $n=199\dots 9$. As $1009=112\cdot 9+1$, so these variants are possible, and number $n$ has to have in the end 112 digits 9. The least value for $n$ is $n=\underset{112}{\underbrace{199\dots 9}}$.