Is there a 6-digit number $\overline{a_1a_2\dots a_6}$ such that the sum of cubes $$\overline{a_1a_2\dots a_6}^3+\overline{a_6a_5\dots a_1}^3$$ is divisible by $10^6$?
We have the usual sum of cubes factorization $$\overline{a_1a_2\dots a_6}^3+\overline{a_6a_5\dots a_1}^3=(\overline{a_1a_2\dots a_6}+\overline{a_6a_5\dots a_1})(\overline{a_1a_2\dots a_6}^2+\overline{a_1a_2\dots a_6}\cdot\overline{a_6a_5\dots a_1}+\overline{a_6a_5\dots a_1}^2)$$
If we make the first sum divisible by $10$, then by case checking we find that the second sum cannot be divisible by $10$. But then we cannot make the first sum divisible by $10^6$ either.