Inspired by Alisa Meier

The maximum average could be determined by assuring that the highest scores give a greater weight on the average. That being said, we have to place the nine students with the highest scores on one class each. All the remaining students will be placed in the tenth class.

And voila, the maximum average will be

$\large \frac {100 + 99 + 98 + 97 + 96 + 95 + 94 + 93 + 92 + \large \frac {91+90+89+...+3+2+1}{91}} {10} = \boxed{91}$ .

Suppose that the student scoring $j$ is in a set of size $n_j$ , for $1 \le j \le 100$ . Then the average of the set averages is equal to $AA \; = \; \tfrac{1}{10}\sum_{j=1}^{100} \frac{j}{n_j}$ By the Rearrangement Theorem, for a given collection of set sizes, the value of $AA$ will be maximised when the sequence $n_j^{-1}$ is increasing, and so the sequence $n_j$ is decreasing. This means that the marks of members of each set form a consecutive block of integers, and the lower scoring sets are larger in size.

Thus we only need to consider set arrangements as follows. There exist integers $1 \; = \; a_1 < a_2 < a_3 < \cdots < a_{10} < a_{11} = 101$ such that

• Set $j$ contains those students who scored marks $a_j,a_j+1,...,a_{j+1}-1$ for any $1 \le j \le 10$ ,
• The set sizes $a_2-a_1, a_3-a_2,...,a_{11} - a_{10}$ form a decreasing sequence.

The average of the marks in Set $j$ is $\tfrac12(a_j + a_{j+1}-1)$ , and so the average of averages is $AA \; = \; \frac{1}{10}\sum_{j=1}^{10} \tfrac12\big(a_j + a_{j+1} - 1\big) \; = \; \tfrac{1}{10}\big[\tfrac12a_1 + a_2 + a_3 + \cdots + a_{10} + \tfrac12a_{11} - 5\big] \; = \; \tfrac{1}{10}\big[46 + a_2 + a_3 + \cdots + a_{10}\big]$ From this it is clear that $AA$ is maximised when $a_j = 90+j$ for $2 \le j \le 10$ . The headmaster creates a set of $91$ pupils (those who scored $1$ to $91$ ), and then $9$ sets with $1$ pupil in each. Thus the maximum possible value of $AA$ is $\tfrac{1}{10}\big[46 + 92 + 93 +\cdots + 100\big] \; = \; \boxed{91}$