8. C1 (PUR) There are 10001 students at a university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold: (i) Each pair of students are in exactly one club. (ii) For each student and each society, the student is in exactly one club of the society. (iii) Each club has an odd number of students. In addition, a club with $2 m+1$ students ( $m$ is a positive integer) is in exactly $m$ societies. Find all possible values of $k$.

The above text is already in English. Here is the text with the same formatting and line breaks:

8. C1 (PUR) There are 10001 students at a university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold: (i) Each pair of students are in exactly one club. (ii) For each student and each society, the student is in exactly one club of the society. (iii) Each club has an odd number of students. In addition, a club with $2 m+1$ students ( $m$ is a positive integer) is in exactly $m$ societies. Find all possible values of $k$.
8. Let us write \( n = 10001 \). Denote by \(\mathcal{T}\) the set of ordered triples \((a, C, \mathcal{S})\), where \(a\) is a student, \(C\) a club, and \(\mathcal{S}\) a society such that \(a \in C\) and \(C \in \mathcal{S}\). We shall count \(|\mathcal{T}|\) in two different ways. Fix a student \(a\) and a society \(\mathcal{S}\). By (ii), there is a unique club \(C\) such that \((a, C, \mathcal{S}) \in \mathcal{T}\). Since the ordered pair \((a, \mathcal{S})\) can be chosen in \(n k\) ways, we have that \(|\mathcal{T}| = n k\). Now fix a club \(C\). By (iii), \(C\) is in exactly \((|C| - 1) / 2\) societies, so there are \(|C|(|C| - 1) / 2\) triples from \(\mathcal{T}\) with second coordinate \(C\). If \(\mathcal{C}\) is the set of all clubs, we obtain \(|\mathcal{T}| = \sum_{C \in \mathcal{C}} \frac{|C|(|C| - 1)}{2}\). But we also conclude from (i) that 
\[ \sum_{C \in \mathcal{C}} \frac{|C|(|C| - 1)}{2} = \frac{n(n - 1)}{2} \]
Therefore \(\frac{n(n - 1)}{2} = n k\), i.e., \(k = \frac{n - 1}{2} = 5000\). On the other hand, for \(k = \frac{n - 1}{2}\) there is a desired configuration with only one club \(C\) that contains all students and \(k\) identical societies with only one element (the club \(C\)). It is easy to verify that (i)-(iii) hold.