99 consecutive positive integers whose sum is a perfect cube.

The sum of the first $n$ consecutive positive integers is the $n$ th triangle number

$T_n=1+2+3+\ldots +n=\frac{n(n+1)}{2}$ .

Thus, the sum of the $99$ consecutive integers starting at $n+1$ is given by

$T_{n+99}-T_n=\frac{(n+99)(n+100)}{2}-\frac{n(n+1)}{2}=\frac{n^2+199n+9900}{2}-\frac{n^2+n}{2}=\frac{198n+9900}{2}=99n+4950=99(n+50)=m^3$

for some positive integer $m$ . Since $99=3^2 \cdot 11$ , the smallest $n$ that guarantees such an $m$ must be of the form $n=3 \cdot 11^2 \cdot k^3$ , for some $k$ . Since we want the smallest such $n$ , we can choose $k=1$ . Thus, $n=3 \cdot 11^2=313$ , so the smallest integer in the sum is $314$