, 2021

Since the average of the two smallest integers in $S$ is 5, their sum is $2 \cdot 5 = 10$.

Since the average of the two largest integers in $S$ is 22, their sum is $2 \cdot 22 = 44$.

Suppose that the other five integers in the set $S$ are $p < q < r < t < u$. (Note that the integers in $S$ are all distinct.)
The average of the nine integers in $S$ is thus equal to $\dfrac{10+44+p+q+r+t+u}{9}$ which equals $6 + \dfrac{p+q+r+t+u}{9}$.

We would like this average to be as large as possible.

To make this average as large as possible, we want $\dfrac{p+q+r+t+u}{9}$ to be as large as possible, which means that we want $p+q+r+t+u$ to be as large as possible.

What is the maximum possible value of $u$?

Let $x$ and $y$ be the two largest integers in $S$, with $x<y$. Since $x$ and $y$ are the two largest integers, then $u<x<y$.

Since $x+y=44$ and $x<y$ and $x$ and $y$ are integers, then $x \leq 21$.

For $u$ to be as large as possible (which will allow us to make $p$, $q$, $r$, $t$ as large as possible), we set $x=21$.

In this case, we can have $u = 20$.

To make $p$, $q$, $r$, $t$ as large as possible, we can take $p=16$, $q=17$, $r=18$, $t=19$.
Here, $p+q+r+t+u=90$.

If $x < 21$, then $p+q+r+t+u$ will be smaller and so not give the maximum possible value.

This means that the maximum possible average of the integers in $S$ is $6 + \dfrac{90}{9} = 16$.