1. A university has 10001 students, some of whom join and form several clubs (a student can belong to different clubs), and some clubs join together to form several associations (a club can belong to different associations). It is known that there are $k$ associations in total. Assume the following conditions are met:
(1) Every pair of students (i.e., any two students) belong to exactly one club;
(2) For each student and each association, the student belongs to exactly one club of that association;
(3) Each club has an odd number of students, and a club with $2m$ +1 students belongs to exactly $m$ associations, where $m$ is a positive integer.
Find all possible values of $k$.
1. Replace 10001 with $n$, and use two methods to calculate the number of ordered triples $(a, R, S)$, where $a, R, S$ represent a student, a club, and a society, respectively, and satisfy $a \in R, R \in S$. We call such triples "acceptable".

Fix a student $a$ and a society $S$. By condition (2), there is a unique club $R$ such that $(a, R, S)$ is an acceptable triple. Since there are $n k$ ways to choose the ordered pair $(a, S)$, there are $n k$ acceptable triples.

Fix a club $R$, and let the number of students in $R$ be $|R|$. By condition (3), $R$ belongs to exactly $\frac{|R|-1}{2}$ societies. Therefore, the number of acceptable triples containing club $R$ is $\frac{|R|(|R|-1)}{2}$. If $M$ is the set of all clubs, then the total number of acceptable triples is $\sum_{R \in M} \frac{|R|(|R|-1)}{2}$. Therefore,
$$
n k=\sum_{R \in M} \frac{|R|(|R|-1)}{2} .
$$

Since $\sum_{R \in M} \frac{|R|(|R|-1)}{2}=\sum_{R \in M} \mathrm{C}_{|R|}^{2}$, by condition (1), we have $\sum_{k \in M} \mathrm{C}_{|k|}^{2}=\mathrm{C}_{n}^{2}$. Hence, $n k=\frac{n(n-1)}{2}$, which implies $k=\frac{n-1}{2}$.
When $n=10001$, $k=5000$.