The Political Chameleons

The highest number of red chameleons will result if they are distributed three per group into as many groups as possible. This is $\frac{249}{3}=83$ groups, or $83\times 5=415$ chameleons. The lowest number of red chameleons will be found when the blue ones do the same. $251$ is not divisible by $3$ , but the two remaining chameleons can't change the result no matter where they are placed. So the largest number of blue chameleons is identical, $415$ , and the smallest number of red is $500-415=85$ corresponding to only $17$ groups. (Quite a difference this kind of redistricting can make.) $n$ has to be divisible by $5$ , since each group ends up being a single color, so we are looking for the number of multiples of $5$ from $85$ to $415$ inclusive, as all the various numbers of groups can be realized (and with a higher probability than the extremes). Or we can just count the number of different red groups, from $17$ to $83$ inclusive. This is $83-17+1=67$ .