In a massive school which has $m$ students, and each student took at least one subject. Let $p$ be an odd prime. Given that:

(i) each student took at most $p+1$ subjects. \\
(ii) each subject is taken by at most $p$ students. \\
(iii) any pair of students has at least $1$ subject in common. \\

Find the maximum possible value of $m$.
1. **Graph Representation**:
   - Let \( G \) be a bipartite graph with vertex sets \( M \) and \( N \), where \( M \) represents the \( m \) students and \( N \) represents the \( n \) subjects.
   - An edge between a student \( A \in M \) and a subject \( C \in N \) indicates that student \( A \) is taking subject \( C \).

2. **Counting Common Subjects**:
   - We need to count the number of triples \((A, B, C)\) such that \( A, B \in M \), \( C \in N \), and both \( A \) and \( B \) are connected to \( C \) in \( G \).
   - Each pair of students has at least one subject in common, so the number of such triples is \( \binom{m}{2} \).

3. **Degree Constraints**:
   - Each subject is taken by at most \( p \) students, so the degree of any vertex in \( N \) is at most \( p \).
   - The number of triples \((A, B, C)\) can also be counted by summing over all subjects the number of ways to choose 2 students from those taking the subject:
     \[
     \sum_{C \in N} \binom{\deg(C)}{2} \leq n \cdot \binom{p}{2}
     \]
   - Since \(\binom{p}{2} = \frac{p(p-1)}{2}\), we have:
     \[
     \sum_{C \in N} \binom{\deg(C)}{2} \leq n \cdot \frac{p(p-1)}{2}
     \]

4. **Edge Counting**:
   - Let \( e \) be the number of edges in \( G \).
   - Each student takes at most \( p+1 \) subjects, so:
     \[
     e \leq (p+1)m
     \]
   - Each subject is taken by at most \( p \) students, so:
     \[
     e \geq p \cdot n
     \]

5. **Combining Inequalities**:
   - From the edge counting inequalities, we get:
     \[
     p \cdot n \leq e \leq (p+1)m
     \]
   - Combining this with the triple counting inequality:
     \[
     \binom{m}{2} \leq n \cdot \frac{p(p-1)}{2}
     \]
   - Simplifying, we get:
     \[
     \frac{m(m-1)}{2} \leq n \cdot \frac{p(p-1)}{2}
     \]
     \[
     m(m-1) \leq n \cdot p(p-1)
     \]

6. **Maximizing \( m \)**:
   - Using \( e \geq p \cdot n \) and \( e \leq (p+1)m \), we get:
     \[
     p \cdot n \leq (p+1)m
     \]
     \[
     n \leq \frac{(p+1)m}{p}
     \]
   - Substituting \( n \) in the inequality \( m(m-1) \leq n \cdot p(p-1) \):
     \[
     m(m-1) \leq \frac{(p+1)m}{p} \cdot p(p-1)
     \]
     \[
     m(m-1) \leq (p+1)m(p-1)
     \]
     \[
     m-1 \leq (p+1)(p-1)
     \]
     \[
     m-1 \leq p^2 - 1
     \]
     \[
     m \leq p^2
     \]

7. **Construction**:
   - To achieve \( m = p^2 \), we need \( n = p(p+1) \).
   - Assign students and subjects such that each student takes exactly \( p+1 \) subjects and each subject is taken by exactly \( p \) students, ensuring that any pair of students shares at least one subject.

The final answer is \( \boxed{ p^2 } \)