49th Austrian Mathematical Olympiad, National Competition (Final Round, part 1)

We first show that $1$, $2$, $3$, $4$, $5$ are elements of $M$: As divisors of $2018$, the numbers $1$, $2$ and $1009$ are elements of $M$. Therefore, $2019=2\cdot 1009+1$ and its divisor $3$ are elements of $M$. We now obtain $7=2\cdot 3+1$ and $15=2\cdot 7+1$ and therefore the divisor $5$ of $15$ as elements of $M$. Considering $16=3\cdot 5+1$, we see that $4\in M$.

We now show by induction that $\{1,2,\dots ,2k-1\}\subseteq M$ for $k\ge 1$. This has been shown above for $k\le 3$. Assume that the assertion holds for some $k\ge 3$. Then we only have to verify that $2k$ and $2k+1$ are elements of $M$, too.

It is clear that $2k+1=2\cdot k+1$ is an element of $M$ due to $k\ge 3$. This implies that $(2k{)}^2=(2k-1)(2k+1)+1$ and its divisor $2k$ are elements of $M$. This concludes the proof of the assertion and shows that $M$ consists of all positive integers.