There are $10001$ students at an 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]i.)[/i] Each pair of students are in exactly one club.

[i]ii.)[/i] For each student and each society, the student is in exactly one club of the society.

[i]iii.)[/i] Each club has an odd number of students. In addition, a club with ${2m+1}$ students ($m$ is a positive integer) is 
in exactly $m$ societies.

Find all possible values of $k$.

[i]Proposed by Guihua Gong, Puerto Rico[/i]
1. Define the set \(\mathcal{A} = \{ (s, C, S_0) \mid s \in C, C \in S_0, s \text{ is a student}, C \text{ is a club, and } S_0 \text{ is a society} \}\). According to the problem, each student is in exactly one club of each society, so the size of \(\mathcal{A}\) is \(10001k\).

2. Let \(C_i\) be a particular club, and denote \(|C_i|\) as the number of students in it. Since each club has an odd number of students, let \(|C_i| = 2m_i + 1\) for some positive integer \(m_i\). According to the problem, a club with \(2m_i + 1\) students is in exactly \(m_i\) societies.

3. The total number of elements in \(\mathcal{A}\) can also be calculated by summing over all clubs. For each club \(C_i\), there are \(|C_i|\) students, and each student in \(C_i\) is in exactly one club of each society that \(C_i\) belongs to. Since \(C_i\) belongs to \(m_i\) societies, the contribution of \(C_i\) to \(\mathcal{A}\) is \(|C_i| \cdot m_i\).

4. Therefore, we have:
   \[
   |\mathcal{A}| = \sum_{i=1}^{n} |C_i| \cdot m_i
   \]
   Given that \(|C_i| = 2m_i + 1\), we can rewrite \(m_i\) as \(\frac{|C_i| - 1}{2}\). Substituting this in, we get:
   \[
   |\mathcal{A}| = \sum_{i=1}^{n} |C_i| \cdot \frac{|C_i| - 1}{2}
   \]

5. From condition (i), each pair of students is in exactly one club. The total number of pairs of students is \(\binom{10001}{2}\). This must equal the sum of the number of pairs within each club:
   \[
   \binom{10001}{2} = \sum_{i=1}^{n} \binom{|C_i|}{2}
   \]
   Simplifying, we get:
   \[
   \frac{10001 \cdot 10000}{2} = \sum_{i=1}^{n} \frac{|C_i|(|C_i| - 1)}{2}
   \]

6. Combining the two expressions for \(|\mathcal{A}|\), we have:
   \[
   10001k = \sum_{i=1}^{n} |C_i| \cdot \frac{|C_i| - 1}{2}
   \]
   and
   \[
   \frac{10001 \cdot 10000}{2} = \sum_{i=1}^{n} \frac{|C_i|(|C_i| - 1)}{2}
   \]

7. Equating the two sums, we get:
   \[
   10001k = \frac{10001 \cdot 10000}{2}
   \]
   Solving for \(k\), we find:
   \[
   k = \frac{10001 \cdot 10000}{2 \cdot 10001} = \frac{10000}{2} = 5000
   \]

The final answer is \(\boxed{5000}\).