There are $ n$  students; each student knows exactly $d $  girl students and $d $ boy students ("knowing" is a symmetric relation). Find all pairs $ (n,d) $  of integers .
1. **Initial Assumptions and Definitions:**
   - Let \( n \) be the total number of students.
   - Each student knows exactly \( d \) girl students and \( d \) boy students.
   - Knowing is a symmetric relation, meaning if student \( A \) knows student \( B \), then student \( B \) knows student \( A \).

2. **Special Case \( d = 0 \):**
   - If \( d = 0 \), then no student knows any other student. This condition is trivially satisfied for any \( n \).

3. **General Case \( d \geq 1 \):**
   - Consider the graph \( F \) with vertices divided into two sets: \( B \) (boys) and \( G \) (girls).
   - Let \( |B| = b \) and \( |G| = g \), where \( b + g = n \).

4. **Counting Edges:**
   - Each boy knows \( d \) girls, and each girl knows \( d \) boys.
   - The total number of edges between boys and girls is \( bd \) (from the boys' perspective) and \( gd \) (from the girls' perspective).
   - Since these counts must be equal, we have:
     \[
     bd = gd
     \]
   - Simplifying, we get:
     \[
     b = g
     \]
   - Therefore, \( b = g = \frac{n}{2} \), implying \( n \) must be even.

5. **Feasibility Condition:**
   - For the structure to be realizable, \( n \geq 2d \) must hold because each student knows \( d \) students of each gender.
   - Thus, \( n \) must be at least \( 2d \).

6. **Constructing the Graph:**
   - Index the boys as \( B = \{b_0, b_1, \ldots, b_{\frac{n}{2} - 1}\} \) and the girls as \( G = \{g_0, g_1, \ldots, g_{\frac{n}{2} - 1}\} \) in \( \mathbb{Z}_{\frac{n}{2}} \).
   - Create edges such that each boy \( b_i \) is connected to girls \( g_i, g_{i+1}, \ldots, g_{i+d-1} \) (indices taken modulo \( \frac{n}{2} \)).

7. **Regularity Condition:**
   - The subgraphs \( B \) and \( G \), each of order \( \frac{n}{2} \), need to be \( d \)-regular.
   - A \( d \)-regular graph on \( \frac{n}{2} \) vertices is possible if and only if \( d \leq \frac{n}{2} - 1 \) and \( d \left(\frac{n}{2}\right) \) is even.

8. **Conclusion:**
   - The pairs \( (n, d) \) that satisfy the conditions are those where:
     - \( n \) is even,
     - \( \frac{n}{2} \geq d + 1 \),
     - \( d \left(\frac{n}{2}\right) \) is even.

The final answer is \( \boxed{ (n, d) } \) such that \( n \) is even, \( \frac{n}{2} \geq d + 1 \), and \( d \left(\frac{n}{2}\right) \) is even.