The country has $n \ge 3$ airports, some pairs of which are connected by bidirectional flights. Every day, the government closes the airport with the strictly highest number of flights going out of it. What is the maximum number of days this can continue?

[i]Proposed by Fedir Yudin[/i]
1. **Base Case for \( n = 3 \):**
   - For \( n = 3 \), consider a graph with 3 vertices and 2 edges. The possible configurations are:
     - A triangle (each vertex connected to the other two).
     - A line (one vertex connected to the other two, and those two are not connected to each other).
   - In both cases, removing the vertex with the highest degree (2) leaves a graph with only one edge, and no further removals are possible. Thus, the maximum number of days is 1.

2. **General Case for \( n \ge 4 \):**
   - We need to construct a graph such that the removal process can continue for \( n - 3 \) days.
   - Label the vertices \( v_1, v_2, \ldots, v_n \).
   - Connect \( v_1 \) to \( v_2, v_3, \ldots, v_{n-1} \).
   - Connect \( v_n \) to \( v_2, v_3, \ldots, v_{n-2} \).
   - Continue this pattern until \( v_5 \) is connected to \( v_2, v_3 \).

3. **Verification of the Construction:**
   - Initially, \( v_1 \) has the highest degree and is removed first.
   - After removing \( v_1 \), \( v_n \) has the highest degree and is removed next.
   - This pattern continues with \( v_{n-1}, v_{n-2}, \ldots \) being removed in sequence.
   - Each removal reduces the graph to a smaller graph with the same structure, allowing the process to continue.

4. **Upper Bound for \( n = 4 \):**
   - Suppose we could make two moves. The first move must delete a vertex with degree at least 2.
   - If the first vertex deleted has degree 2, the next deletion must have a vertex with degree 1, which is not possible since the adjacent vertex would also have degree 1.
   - Therefore, the first vertex deleted must have degree 3. After its deletion, all remaining vertices have degrees at most 1, preventing further deletions.

5. **General Upper Bound for \( n > 4 \):**
   - After each move, the problem reduces to a smaller graph with \( n - 1 \) vertices.
   - Thus, we can perform at most \( n - 4 \) moves, leading to a total of \( n - 3 \) moves.

\(\blacksquare\)

The final answer is \( \boxed{ n - 3 } \) for \( n \ge 4 \) and 1 for \( n = 3 \).