The maximum possible $n$ is $200$.

Let us assume that there is a contestant, say $X$, who knows $201$ other contestants and let those $201$ contestants make up a set $S$. There must exist contestants who know exactly $1$, $2$, $\dots$, $200$ other contestants.

We will say that the contestant *has degree* $m$ if he knows exactly $m$ other contestants.

Each element in $S$ has degree at most $99$. Namely, he cannot know anyone from $S$ since there are no three contestants who all know each other (aside from $X$ and $S$, there are only $98$ contestants).

We conclude that contestants from $S$ have at most $99$ distinct degrees. Simultaneously, there are $98$ contestants not in $S$ and different from $X$, so they can have at most $98$ distinct degrees.

This shows that there are at most $1 + 99 + 98 = 198 < 201$ distinct degrees, and therefore it is impossible that there are contestants who know exactly $1$, $2$, $\dots$, $201$ other contestants. Hence, there is no contestant who knows exactly $201$ other contestants.

Let us now show that $n = 200$ is possible. Denote $100$ contestants as $A_1, A_2, \dots, A_{100}$ and call them A-contestants, and the remaining $200$ as $B_1, B_2, \dots, B_{200}$ and call them B-contestants.

For each $i \in \{1, 2, \dots, 100\}$ and each $j \in \{1, 2, \dots, 200\}$, such that $i \le j$, let $A_i$ and $B_j$ be acquainted. All the other pairs of contestants are unfamiliar with each other. We claim that this example has $n = 200$, and that all the conditions of the problem are satisfied.

Namely, there are no three contestants who all know each other, because no two A-contestants know each other and no two B-contestants know each other.

Moreso, for $i \in \{1, 2, \dots, 100\}$, contestant $A_i$ knows contestants $B_i, B_{i+1}, \dots, B_{200}$ and only them. Hence, he has exactly $201 - i$ acquaintances, which means that there are contestants knowing exactly $200$, $199$, $\dots$, $101$ other contestants. Analogously, for $j \in \{1, 2, \dots, 100\}$, contestant $B_j$ knows $A_1, \dots, A_{j-1}, A_j$ and only them. This implies that he has exactly $j$ acquaintances, which means that there are contestants knowing exactly $1$, $2$, $\dots$, $100$ other contestants.