Let $n\geq 2$ be an integer. Alice and Bob play a game concerning a country made of $n$ islands. Exactly two of those $n$ islands have a factory. Initially there is no bridge in the country. Alice and Bob take turns in the following way. In each turn, the player must build a bridge between two different islands $I_1$ and $I_2$ such that:
$\bullet$ $I_1$ and $I_2$ are not already connected by a bridge.
$\bullet$ at least one of the two islands $I_1$ and $I_2$ is connected by a series of bridges to an island with a factory (or has a factory itself). (Indeed, access to a factory is needed for the construction.)
As soon as a player builds a bridge that makes it possible to go from one factory to the other, this player loses the game. (Indeed, it triggers an industrial battle between both factories.) If Alice starts, then determine (for each $n\geq  2$) who has a winning strategy.
([i]Note:[/i] It is allowed to construct a bridge passing above another bridge.)

1. **Restate the problem in graph theory terms:**
   - We have a graph with \( n \) vertices (islands) and initially no edges (bridges).
   - Two specific vertices have factories.
   - Players take turns adding edges under certain conditions.
   - The player who connects the two factory vertices loses.

2. **Analyze the game for different values of \( n \mod 4 \):**
   - We need to determine the winning strategy based on the value of \( n \mod 4 \).

3. **Case 1: \( n \) is even:**
   - If \( n \) is even, Bob can mirror Alice's moves.
   - Divide the vertices into two groups of \( n/2 \) vertices each.
   - Alice and Bob can each build bridges within their respective groups.
   - Eventually, Alice will be forced to connect the two groups, causing her to lose.
   - Therefore, Bob wins when \( n \) is even.

4. **Case 2: \( n \equiv 1 \pmod 4 \):**
   - Consider the sum of the edges in the two complete subgraphs formed by Alice and Bob.
   - The sum of the edges in a complete graph with \( k \) vertices is \( \binom{k}{2} \).
   - We need to show that the sum of the edges in the two subgraphs is even.
   - If \( x + y \equiv 1 \pmod 4 \), then \( \binom{x}{2} + \binom{y}{2} \) is even.
   - Therefore, Alice will be forced to make the losing move, and Bob wins.

5. **Case 3: \( n \equiv 3 \pmod 4 \):**
   - Similar to the previous case, consider the sum of the edges in the two complete subgraphs.
   - If \( x + y \equiv 3 \pmod 4 \), then \( \binom{x}{2} + \binom{y}{2} \) is odd.
   - Therefore, Bob will be forced to make the losing move, and Alice wins.

6. **Summarize the results:**
   - Alice wins when \( n \equiv 3 \pmod 4 \).
   - Bob wins otherwise.

The final answer is Alice wins when \( \boxed{ n \equiv 3 \pmod 4 } \), and Bob wins otherwise.