South African Mathematics Olympiad Third Round

$$\begin{array}{rl}\frac{n}{3}+\frac{n^2}{2}+\frac{n^3}{6}& =\frac{2n+3n^2+n^3}{6}\\ & =\frac{n(2+3n+n^2)}{6}\\ & =\frac{n(n+1)(n+2)}{6}\end{array}$$ 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.