The minimal sum for $n$ numbers in the list, is $\frac{100 + \frac{n(n+1)}{2}}{n+1} = \frac{200+n^2+n}{2n+2}$, which is minimized for $n = 10\sqrt{2}-1$, giving us $10\sqrt{2}-1$ in return. This means that every number $\ge 14$ could be reached in theory. Now the upperbound: The maximal sum clearly is $99$. In theory, we can reach every number between $13$ and $100$, we just need to find configurations. This is pretty trivial to be honest. If the sum is too small, you can replace with larger one etc.

As configuration we can use next sets If $14 \leq n\leq 50$ then set $\{n-13,n-12,n-11,...,n-2,2n-10,100 \}$ For $51 \leq n \leq 99$ we can use set $\{2n-100,100\}$

If the maximum is 100 then the average cannot be greater than 100 but since the numbers are never said to be different then we can have a set of several 100 and their average would be 100. The larger the set, the smaller the average can be. For example, if we have a set of 100 elements, then we can divide any multiple of 100 into 99 and we will then have the numbers from 2 to 100. like this: The average of $\{1,1,1,1,...,2,100\}$ is 2 The average of $\{2,2,2,2,...,4,100\}$ is 3 we cannot make the average 1 because there will always be 99 left over For example, in a set of 1000 we have that 999 ones plus 100 is equal to 1099 and for the average it is divided by 1000