For how many positive integers $n$ is the sum

$$
(-n)^{3}+(-n+1)^{3}+\cdots+(n-2)^{3}+(n-1)^{3}+n^{3}+(n+1)^{3}
$$

less than $3129 ?$
For every real number $a,(-a)^{3}=-a^{3}$ and so $(-a)^{3}+a^{3}=0$.

Therefore,

$$
(-n)^{3}+(-n+1)^{3}+\cdots+(n-2)^{3}+(n-1)^{3}+n^{3}+(n+1)^{3}
$$

which equals

$$
\left((-n)^{3}+n^{3}\right)+\left((-n+1)^{3}+(n-1)^{3}\right)+\cdots+\left((-1)^{3}+1^{3}\right)+0^{3}+(n+1)^{3}
$$

is equal to $(n+1)^{3}$.

Since $14^{3}=2744$ and $15^{3}=3375$ and $n$ is an integer, then $(n+1)^{3}<3129$ exactly when $n+1 \leq 14$.

There are 13 positive integers $n$ that satisfy this condition.

ANSWER: 13