HMMT November

Solution:
I claim that $a_{i}=3i$ for all $i$. We can conjecture that the sequence should just be the positive multiples of three because the natural numbers satisfy the property that the square of their sum is the sum of their cubes, and prove this by induction. At $i=1$, we have that $3 a_{i}^{2}=a_{i}^{3}$, so $a_{i}=3$. Now assuming this holds for $i=m$, we see that
$$\begin{aligned} 3\left(\sum_{i=1}^{m+1} a_{i}\right)^{2} & =3\left(a_{m+1}+\sum_{i=1}^{m} a_{i}\right)^{2} \\ & =3 a_{m+1}^{2}+\sum_{i=1}^{m} a_{i}^{3}+6 a_{m+1} \sum_{i=1}^{m} a_{i} \\ & =3 a_{m+1}^{2}+\sum_{i=1}^{m} a_{i}^{3}+6 a_{m+1} \cdot 3\left(\frac{m(m+1)}{2}\right) \\ & =\sum_{i=1}^{m+1} a_{i}^{3} \end{aligned}$$
Therefore,
$$\begin{aligned} a_{m+1}^{3} & =3 a_{m+1}^{2}+a_{m+1}\left(9 m^{2}+9 m\right) \\ 0 & =a_{m+1}^{2}-3 a_{m+1}-\left(9 m^{2}+9 m\right) \\ 0 & =\left(a_{m+1}-(3 m+3)\right)\left(a_{m+1}+3 m\right) \end{aligned}$$
and because the sequence is positive, $a_{m+1}=3 m+3$, which completes the induction. Then $a_{1337}=1337 \cdot 3=4011$.