We claim that for any odd $n, a_{n}=n$. The proof is by induction. To get the base cases $n=1$, 3, we compute $a_{1}=1, a_{2}=\left\lfloor 2^{3} / 1\right\rfloor=8, a_{3}=\left\lfloor 3^{3} / 8\right\rfloor=3$. And if the claim holds for odd $n \geq 3$, then $a_{n+1}=\left\lfloor(n+1)^{3} / n\right\rfloor=n^{2}+3 n+3$, so $a_{n+2}=\left\lfloor(n+2)^{3} /\left(n^{2}+3 n+3\right)\right\rfloor=\left\lfloor\left(n^{3}+6 n^{2}+12 n+8\right) /\left(n^{2}+3 n+2\right)\right\rfloor=\left\lfloor n+2+\frac{n^{2}+3 n+2}{n^{2}+3 n+3}\right\rfloor=n+2$. So the claim holds, and in particular, $a_{999}=999$.