$$\begin{aligned} \frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6} &= \frac{2n + 3n^2 + n^3}{6} \\ &= \frac{n(2 + 3n + n^2)}{6} \\ &= \frac{n(n + 1)(n + 2)}{6} \end{aligned}$$
Since $n$, $n + 1$, $n + 2$ are three consecutive integers, at least one is divisible by $2$ and one is divisible by $3$. Hence $n(n + 1)(n + 2)$ is divisible by $6$ and therefore $\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$ must be an integer.