The only possible (and hence the maximum) $n$ is $448$.

Indeed, by the friendship theorem, there must be one person who knows everyone else. We give a proof as follows.

WLOG, assume $A$ and $B$ do not know each other. If $X$ knows $A$, then $X$ and $B$ know a person $Y$ in common. Note that the function mapping $X$ to $Y$ is well-defined, and it is a bijection between the set of persons known by $A$ and the set of persons known by $B$. Hence any two persons who do not know each other know the same number of persons.

Let $S_1$ be the set of all persons who know exactly $k$ persons, where $k$ is chosen such that $|S_1| \ge 1$. Let $S_2$ be the complement of $S_1$. Note that each person in $S_2$ must know every person in $S_1$. Otherwise, this person must know exactly $k$ persons from the above observation, and hence this person should belong to $S_1$ by definition.

If $|S_2| \ge 2$, then we choose two persons $X$ and $Y$ in $S_2$ arbitrarily. Since they only know one person in common, we must have $|S_1| = 1$. In this case the person in $S_1$ knows everyone.

If $|S_2| = 1$, then the person in $S_2$ knows everyone.

If $|S_2| = 0$, then everyone knows exactly $k$ persons. On the one hand, there are $\binom{451}{2}$ pairs of persons in this group. On the other hand, there are $\binom{k}{2}$ pairs of persons knowing a particular person in common. Therefore, we have
$$451 \binom{k}{2} = \binom{451}{2}.$$

But this implies $k(k-1) = 450$, which has no solution. (In this case, we make use of the fact that $451$ is not of the form $k^2 - k + 1$. But the friendship theorem does apply to any number of persons.)

Therefore, we have proven there exists a person $A$ who knows everyone. For every other person $B$, in order that $A$ and $B$ know exactly one person in common, $B$ must know exactly one person apart from $A$. Thus, the other persons can be paired up such that only the two persons in the same pair know each other. Hence, the only possible $n$ is $450 - 2 = 448$.