Croatian Junior Mathematical Olympiad

We show that $n$ can be $0,1,2,3,\dots ,98$. The following contains the description of a situation in which exactly $n$ people are present after the last chime (for $n=0,1,2,\dots ,98$): For $n>0$, we can divide all the people at the party into two groups, A and B. Let A contain $n$ people and assume every one of them is acquainted with all the other people at the party. Let B contain the remaining $100-n$ people, none of whom are acquainted amongst themselves, but all of whom are acquainted with all the people in group A (thus, every person in group B has $n$ acquaintances). All the people from group B will obviously leave the party after the $(n+1{)}^{\text{th}}$ sounding of the gong. After that, only people from group A will remain, and each of them will have exactly $n-1$ acquaintances. As the gong has already been sounded $n$ times, this means none of them will leave until the end of the party. The value $n=0$ is attained, for example, when all the party-goers know each other (so they all leave after the $100$th chime). A simple observation shows that at least one person must leave at some moment: this is the person with the fewest acquaintances. This implies that $n=100$ cannot be attained. Finally, let us prove that $n=99$ is not possible. Assume the contrary, i.e. that exactly one person will leave the party before it ends; call this person $X$. As the first one to leave, this is obviously the person with the fewest acquaintances. Since none of the remaining people leave after $X$, we conclude that all of them must be acquainted with $X$ - if another person $Y$ is not acquainted with $X$, then the number of people they know would not change with $X$ leaving, so they would have to leave at some point as well. This shows that $X$ has $99$ acquaintances, which contradicts the assumption that $X$ has the fewest acquaintances.