2. For which natural numbers $n$ will there exist a natural number $k$ such that the number $2 k^{2}+k+2018$ is divisible by $n$ ! (as usual, $n$ ! denotes the product of all natural numbers not exceeding $n$, for example, $4!=1 \cdot 2 \cdot 3 \cdot 4) ?$
Answer: $n=1,2,3,4$.

Solution. Consider the remainders of the number $2 k^{2}+k$ when divided by 5: $2 \cdot 0^{2}+0=0$ gives a remainder of $0, 2 \cdot 1^{2}+1=3$ gives a remainder of 3, $2 \cdot 2^{2}+2=10$ gives a remainder of $0, 2 \cdot 3^{2}+3=21$ gives a remainder of 1, and $2 \cdot 4^{2}+4=36$ gives a remainder of 1. Therefore, the remainder of the number $2 k^{2}+k$ when divided by 5 cannot be 2. Thus, $2 k^{2}+k+2018$ cannot be divisible by 5. Since for $n \geqslant 5$ the number $n!$ is divisible by 5, numbers $n \geqslant 5$ are definitely not suitable. For the remaining natural numbers $n$, it is not difficult to find such a $k$, for example, $k=14$ works, since $2 \cdot 14^{2}+14+2018=2424=$ $24 \cdot 101$ is divisible by $4!$.