Solution:

Let $S(i)$ be the favorite seat of the $i$th person, counting from the right. Let $P(n)$ be the probability that at least $n$ people get to sit. At least $n$ people sit if and only if $S(1) \geq n, S(2) \geq n-1, \ldots, S(n) \geq 1$. This has probability:
$$P(n) = \frac{100-(n-1)}{100} \cdot \frac{100-(n-2)}{100} \cdots \frac{100}{100} = \frac{100!}{(100-n)! \cdot 100^{n}}$$
The probability, $Q(n)$, that exactly $n$ people sit is
$$Q(n) = P(n) - P(n+1) = \frac{100!}{(100-n)! \cdot 100^{n}} - \frac{100!}{(99-n)! \cdot 100^{n+1}} = \frac{100! \cdot n}{(100-n)! \cdot 100^{n+1}}$$
Now,
$$\frac{Q(n)}{Q(n-1)} = \frac{100! \cdot n}{(100-n)! \cdot 100^{n+1}} \cdot \frac{(101-n)! \cdot 100^{n}}{100! \cdot (n-1)} = \frac{n(101-n)}{100(n-1)} = \frac{101 n - n^{2}}{100 n - 100},$$
which is greater than $1$ exactly when $n^{2} - n - 100 < 0$, that is, for $n \leq 10$. Therefore, the maximum value of $Q(n)$ occurs for $n = 10$.