In a party of $2020$ people, some pairs of people are friends. We say that a given person's [i]popularity[/i] is the size of the largest group of people in the party containing them with the property that every pair of people in that group is friends. A person has popularity number $1$ if they have no friends. What is the largest possible number of distinct popularities in the party?


[i]2021 CCA Math Bonanza Tiebreaker Round #3[/i]
1. Let \( k \) be the largest popularity number in the party. This means there exists a group of \( k \) people who are all mutual friends.
2. If \( k \) is the largest popularity number, then there are at most \( k \) distinct popularity numbers. This is because the largest clique (a group where every pair of people are friends) determines the maximum popularity number.
3. Consider the people \( P_1, P_2, \ldots, P_k \) who are all mutual friends. Each of these people has a popularity number \( k \). Therefore, the number of distinct popularity numbers is at most \( 1 + (2020 - k) = 2021 - k \). This is because the remaining \( 2020 - k \) people can have popularity numbers ranging from 1 to \( k-1 \).
4. To find the maximum number of distinct popularity numbers, we need to maximize the minimum of \( k \) and \( 2021 - k \). This is because the number of distinct popularity numbers is bounded by both \( k \) and \( 2021 - k \).
5. The function \( \min(k, 2021 - k) \) is maximized when \( k = 1010 \). This is because \( \min(k, 2021 - k) \leq 1010 \) for all \( k \).
6. Therefore, the upper bound for the number of distinct popularity numbers is \( 1010 \).
7. To achieve this bound, we can construct a scenario where \( P_{1010}, P_{1011}, \ldots, P_{2020} \) form a group of \( 1011 \) mutual friends. For each \( n \in \{1, 2, \ldots, 1009\} \), let \( P_n \) be friends with \( n \) randomly chosen people from the group of \( 1011 \) mutual friends. This construction ensures that there are \( 1010 \) distinct popularity numbers.

The final answer is \( \boxed{ 1010 } \).