For a given positive integer $n$, try to find all positive integers $k$ such that $2^{n}$ can be expressed as the sum of cubes of $k$ consecutive integers (including 1 integer).
(Feng Yuefeng, problem contributor)
The conclusion of this problem is: when $n$ is a multiple of 3, $k=1,2^{\frac{n+3}{3}}$; when $n$ is not a multiple of 3, $k$ does not exist.
Let
$$
2^{n}=(i+1)^{3}+(i+2)^{3}+\cdots+(i+k)^{3} \text {. }
$$

If $i \geqslant 0$, then let $S_{t}=1^{3}+2^{3}+\cdots+t^{3}\left(t \in \mathbf{N}\right.$, and define $\left.S_{0}=0\right)$, then equation (1) becomes $2^{n}=S_{i+k}-S_{i}$, where $0 \leqslant i0)$, then equation (1) becomes
$$
2^{n}=(-r+1)^{3}+(-r+2)^{3}+\cdots+(-r+k)^{3} \text {. }
$$

Noting that $2^{n}>0$, the positive terms on the right side of the equation outnumber the negative terms, so the equation can be rewritten as
$$
\begin{aligned}
2^{n}= & (1-r)^{3}+(2-r)^{3}+\cdots+0^{3}+1^{3}+\cdots+(r-2)^{3}+(r-1)^{3} \\
& +r^{3}+(r+1)^{3}+\cdots+(-r+k)^{3} \\
= & S_{-r+k}-S_{r-1}=S_{i+k}-S_{-i-1}
\end{aligned}
$$

where $0 \leqslant-i-11$, so $j-i=1$, i.e., $j=i+1$. Therefore, $2^{n}=S_{i+1}-S_{i}=(i+1)^{3}$.
(1) When $n$ is a multiple of 3, from equation (2) we get $i+1=2^{\frac{n}{3}}$, at this time
$$
2^{n}=(i+1)^{3}=\left(2^{\frac{n}{2}}\right)^{3}=\left(2^{\frac{n}{3}}\right)^{3}+\left(2^{\frac{n}{3}}-1\right)^{3}+\cdots+0^{3}+1^{3}+\cdots+\left(1-2^{\frac{n}{3}}\right)^{3},
$$

Thus, $k=1,2 \times 2^{\frac{n}{3}}$ satisfies the condition.
(2) When $n$ is not a multiple of 3, $2^{n}$ is not a cube number, which contradicts equation (2), so $k$ does not exist.
In summary, when $n$ is a multiple of 3, $k=1,2^{\frac{n+3}{3}}$; when $n$ is not a multiple of 3, $k$ does not exist.