9.100 Arrange chairs in a circle, with $n$ people sitting on the chairs, such that when one more person sits down, they will always be sitting next to one of the original $n$ people. The minimum value of $n$ is $\qquad$ .
9.34.

From the problem, we know that after $n$ people sit down, there are at most two empty chairs between any two people. If we can arrange it so that there are exactly two empty chairs between every two people, then $n$ is minimized. In this case, if we number the chairs where people are sitting, we get an arithmetic sequence: $1, 4, 7, \cdots, 100$.
Thus, $100 = 1 + 3(n-1)$. Solving for $n$ gives $n = 34$.