Real life problem: How many finalists can participate per school?

The assignment $(3,3,2,2,2,1,1,1,1)$ is optimal for each of the following three objectives:

• Minimize sum of absolute differences between number of seats and quota
• Minimize maximum of absolute differences between number of seats and quota
• Minimize sum of absolute differences between people per seat and $1338/16$

It seems natural to assign the places roughly proportional to the number of students. Then again, this might easily leave school with $0$ finalists. And we have to deal with rounding. So the next natural thing is to assign $$ lfloor alpha n+betarfloor$$ finalists to a school with $n$ students. Remains to pick $alpha$ and $beta$. Typically, one fixes one of them and adjusts the other (e.g., by trial and error) until $$ sum_{i=1}^9lfloor alpha n_i+betarfloor =16.$$ Different strategies are possible (and some correspond to vote counting methods commonly in use to distribute parliamentary seats according to vote counts):

• Set $beta=0$ and adjust $alpha$. This method tends to favour large "parties" and may not lead to a "seat" for the smallest school here
• Set $alpha=frac{16}{sum n_i}$ and adjust $beta$. This method assigns the naive average as "cost" of a seat and adjusts the offset $beta$ to make the first seats cheaper until the sum is right
• Set $beta=1$ and adjust $alpha$. This is essentially like the first, except that every school gets one seat for free to start with.
• Set $beta=frac12$ and adjust $alpha$. This is somewhere between first and third option
• ...