There are $n$ cities, $2$ airline companies in a country. Between any two cities, there is exactly one $2$-way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of $n$.
1. **Understanding the Problem:**
   We need to find the maximum number of cities \( n \) such that in a graph \( G \) with \( n \) vertices, both \( G \) and its complement \( \overline{G} \) are forests. A forest is a disjoint union of trees, and a tree is an acyclic connected graph.

2. **Graph Properties:**
   - A forest with \( n \) vertices has at most \( n-1 \) edges.
   - The complement of a forest with \( n \) vertices must also be a forest.

3. **Edge Count Analysis:**
   - Let \( e \) be the number of edges in \( G \).
   - Since \( G \) is a forest, \( e \leq n-1 \).
   - The complement \( \overline{G} \) has \( \binom{n}{2} - e \) edges.
   - Since \( \overline{G} \) is also a forest, \( \binom{n}{2} - e \leq n-1 \).

4. **Inequality Setup:**
   \[
   e \leq n-1
   \]
   \[
   \binom{n}{2} - e \leq n-1
   \]
   Substituting \( \binom{n}{2} = \frac{n(n-1)}{2} \):
   \[
   \frac{n(n-1)}{2} - e \leq n-1
   \]
   Rearrange to find \( e \):
   \[
   \frac{n(n-1)}{2} - (n-1) \leq e
   \]
   \[
   \frac{n(n-1) - 2(n-1)}{2} \leq e
   \]
   \[
   \frac{n^2 - 3n + 2}{2} \leq e
   \]

5. **Combining Inequalities:**
   \[
   \frac{n^2 - 3n + 2}{2} \leq e \leq n-1
   \]
   Since \( e \) must be an integer, we need to find \( n \) such that:
   \[
   \frac{n^2 - 3n + 2}{2} \leq n-1
   \]
   Multiply through by 2:
   \[
   n^2 - 3n + 2 \leq 2n - 2
   \]
   Simplify:
   \[
   n^2 - 5n + 4 \leq 0
   \]
   Factorize:
   \[
   (n-1)(n-4) \leq 0
   \]

6. **Solving the Inequality:**
   The inequality \((n-1)(n-4) \leq 0\) holds for \( 1 \leq n \leq 4 \).

7. **Verification:**
   - For \( n = 4 \), we need to check if both \( G \) and \( \overline{G} \) can be forests.
   - Consider a path graph \( P_4 \) with 4 vertices: \( A-B-C-D \).
   - \( G \) has 3 edges, and \( \overline{G} \) has \( \binom{4}{2} - 3 = 3 \) edges.
   - Both \( G \) and \( \overline{G} \) are forests.

Thus, the maximum value of \( n \) is 4.

The final answer is \( \boxed{4} \)