4 consecutive integers

If the smallest integer is $n$ then the sum of the integers is $4n+6$ and $4n+6 \equiv 2 \mod 4$ , but all perfect cubes are $0,1$ or $3 \pmod{4}$ . So The sum of 4 consecutive integers cannot be a perfect cube.

Let the numbers be $n,n+1,n+2,n+3$ . Note that the sum $4n+6 =2(2n+3)$ has an even factor $2$ and an odd factor $2n+3$ . Since both have no common factor and $2$ is not a perfect cube, it can never be a perfect cube. So $\boxed{No}$ .