Beautiful problem, as any other problem on this contest. Mykhailo cannot do it, two odd numbers cannot be consecutive by parity reasons (otherwise all numbers are odd), therefore only possibility remains when odd and even numbers alternate, however then for $b=2024$, we have $a+c$ should be divisible by $2025$, however by above arguments it is also even, so is at least $2 \cdot 2025 = 4050$, which is impssible since both $a$ and $c$ are less than $2025$.