Problem 1589

1. **Initial Elements in $M$**:
- Given $2018 \in M$.
- By property (2), all positive divisors of $2018$ are also in $M$. The divisors of $2018$ are $1, 2, 1009,$ and $2018$. Therefore, $1, 2, 1009 \in M$.

2. Using Property (3) to Generate New Elements:
- Since $2 \in M$ and $1009 \in M$, we can use property (3) to generate new elements.
- For $k = 2$ and $m = 1009$, we have $2 \cdot 1009 + 1 = 2019 \in M$.

3. Generating More Elements:
- Since $2 \in M$ and $2019 \in M$, we can use property (3) again.
- For $k = 2$ and $m = 2019$, we have $2 \cdot 2019 + 1 = 4039 \in M$.

4. Generating Smaller Elements:
- Since $2 \in M$ and $4 \in M$ (as $4$ is a divisor of $2018$), we can use property (3).
- For $k = 2$ and $m = 4$, we have $2 \cdot 4 + 1 = 9 \in M$.
- Since $9 \in M$ and $3 \mid 9$, by property (2), $3 \in M$.

5. Continuing the Process:
- Since $2 \in M$ and $3 \in M$, we can use property (3).
- For $k = 2$ and $m = 3$, we have $2 \cdot 3 + 1 = 7 \in M$.
- Since $2 \in M$ and $7 \in M$, we can use property (3).
- For $k = 2$ and $m = 7$, we have $2 \cdot 7 + 1 = 15 \in M$.
- Since $15 \in M$ and $5 \mid 15$, by property (2), $5 \in M$.

6. Generalizing for All Odd Integers:
- By choosing $m = 2$ and $n \geq 3$, we obtain $2n + 1 \in M$. This implies that all odd integers greater than $7$ belong to $M$.
- Since $1, 3, 5 \in M$, all odd integers belong to $M$.

7. Generalizing for All Even Integers:
- For $t > 1$, by choosing $k = 2t - 1$ and $m = 2t + 1$, we have $(2t - 1)(2t + 1) + 1 = 4t^2 \in M$.
- Since $2t \mid 4t^2$, by property (2), $2t \in M$ for each $t \geq 2$.

8. Conclusion:
- We have shown that all odd integers and all even integers greater than $2$ belong to $M$.
- Since $2 \in M$, it follows that all positive integers belong to $M$.

$\blacksquare$