The player who goes first, that is, Petro, has a winning strategy! It is as follows: Petro should choose $1012=\frac{2024}{2}$ on his first move. After his move, there are $1011$ even numbers and $1012$ odd numbers available. Now, if at any point Vasyl chose an even number, he would lose, as the product would become divisible by $2024$. So, throughout the game, he can only choose odd numbers. There will be $1011$ odd numbers remaining after Vasyl's move. Similarly, Petro is compelled to pick an odd number as well. Proceeding in this fashion, there will be a point where $1011$ odd numbers are exhausted and only one odd number remains. With respect to parity, this situation will arise during Petro's turn. Hence, Petro picks the last odd number and Vasyl is forced to choose an even number, resulting in Petro's win. $\square$