In the morning, $100$ students study as $50$ groups with two students in each group. In the afternoon, they study again as $50$ groups with two students in each group. No matter how the groups in the morning or groups in the afternoon are established, if it is possible to find $n$ students such that no two of them study together, what is the largest value of $n$?

$ 
\textbf{(A)}\ 42
\qquad\textbf{(B)}\ 38
\qquad\textbf{(C)}\ 34
\qquad\textbf{(D)}\ 25
\qquad\textbf{(E)}\ \text{None of above}
$
1. **Graph Representation**:
   - Consider a graph \( G \) with \( 100 \) vertices, each representing a student.
   - In the morning, students are grouped into \( 50 \) pairs, which we represent as \( 50 \) red edges.
   - In the afternoon, students are again grouped into \( 50 \) pairs, represented as \( 50 \) blue edges.
   - Each vertex (student) is connected by exactly one red edge and one blue edge.

2. **Graph Properties**:
   - Since each vertex has degree 2 (one red edge and one blue edge), the graph \( G \) is a union of disjoint cycles.
   - Each cycle alternates between red and blue edges.

3. **Cycle Analysis**:
   - Each cycle must have an even number of vertices because the edges alternate in color.
   - For a cycle with \( 2k \) vertices, we can select \( k \) vertices such that no two selected vertices are adjacent (i.e., no two selected vertices are connected by an edge).

4. **Selecting Students**:
   - From each cycle of \( 2k \) vertices, we can select \( k \) vertices.
   - Since the graph is a union of disjoint cycles, we can apply this selection process to each cycle independently.

5. **Maximizing the Number of Students**:
   - The total number of vertices (students) is \( 100 \).
   - Since we can select half of the vertices from each cycle, the maximum number of students we can select such that no two of them study together is \( \frac{100}{2} = 50 \).

Conclusion:
The largest value of \( n \) such that no two of the \( n \) students study together in either the morning or the afternoon is \( 50 \).

The final answer is \( \boxed{50} \).