Let $n$ be a positive integer. Given are numbers $n^3,n^3+1,\ldots,n^3+n$. Of them, $a$ are colored red, and $b$ others blue. The sum of the red numbers divides the sum of the blue numbers. Prove that $a$ divides $ b$.
Some small cases. For $n=2$, we have $8,9,10$. the only way is to color $9$ red and $8,10$ blue. For $n=3$ we have $27,28,29,30$. The only ways are to color $28$ red and $27,29$ blue, or $29$ red and $28,30$ blue. In all cases, $a$ divides $b$.
My intuition is that the numbers are in a small enough range that the average of the red numbers must be the same as the blue numbers, which implies that $a$ divides $b$.
You're on the right general track with your intuition, but proving that the averages are the same is going to be tricky in its own right. Here's an outline of a related but different approach to the problem:
Note that the sum of the numbers from $1$ to $n$ is approximately $\frac{n^2}{2}$; in particular, it's less than $n^2$ for all $n\gt 1$. This means that the sum of the red numbers — call it $\sum A$ — is between $an^3$ and $an^3+n^2$, and in particular that it's bracketed between $an^3$ and $(a+1)n^3$. Likewise, the sum of the blue numbers ($\sum B$) is between $bn^3$ and $bn^3+n^2$. Now, what do the multiples of $\sum B$ look like? You know that the ratio of $\sum B$ to $\sum A$ has to be less than $n$ (why?), so you should be able to bracket each of the multiples of $\sum B$, $m\cdot\sum B$, between $mbn^3$ and $(mb+1)n^3$ (why?). Now, do you see why this implies your result?