26 Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$.

Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$.
26. Let $n=1000 x+y$, where $x$ is a positive integer, $y$ is an integer, and $0 \leqslant y \leqslant 999$. According to the problem,
$$x^{3}=1000 x+y$$

From $0 \leqslant y \leqslant 999$, we know
$$1000 x \leqslant x^{3}<1000 x+1000=1000(x+1)$$

Thus,
$$x^{2} \geqslant 1000, x^{3}+1 \leqslant 1000(x+1)$$

Therefore,
$$x^{2} \geqslant 1000, x^{2}-x+1 \leqslant 1000$$

So,
$$32 \leqslant x<33$$

Hence,
$$x=32,$$

Then,
$$y=768$$

Therefore,
$$n=32768$$