A 2-player game is played on $n\geq 3$ points, where no 3 points are collinear. Each move consists of selecting 2 of the points and drawing a new line segment connecting them. The first player to draw a line segment that creates an odd cycle loses. (An odd cycle must have all its vertices among the $n$ points from the start, so the vertices of the cycle cannot be the intersections of the lines drawn.) Find all $n$ such that the player to move first wins.
1. **Case 1: \( n \) is odd**

   - When \( n \) is odd, the first player (Jacob) has a winning strategy. Initially, the graph \( G \) is empty and thus bipartite. Let \( G = A \cup B \) be a bipartition of \( G \).
   - The number of edges in a complete bipartite graph \( K_{|A|, |B|} \) is \( |A| \cdot |B| = |A|(n - |A|) \). Since \( n \) is odd, either \( |A| \) or \( n - |A| \) is even, making \( |A|(n - |A|) \) even.
   - Before Jacob's turn, the number of edges in \( G \) is odd. Therefore, \( G \) is not \( K_{|A|, |B|} \), and Jacob can always add an edge between some \( a \in A \) and \( b \in B \) that has not been drawn, keeping \( G \) bipartite and free of odd cycles.
   - Jacob continues this strategy, ensuring he never creates an odd cycle, and thus he will win.

2. **Case 2: \( n \equiv 2 \pmod{4} \)**

   - When \( n \equiv 2 \pmod{4} \), the second player (David) has a winning strategy. On his first turn, David draws a random edge between two vertices \( v_1 \) and \( v_2 \).
   - If Jacob connects \( v_1 \) or \( v_2 \) to another vertex \( v_3 \), David connects \( v_3 \) to another vertex \( v_4 \). If Jacob connects two other vertices \( v_3 \) and \( v_4 \), David connects two other vertices \( v_5 \) and \( v_6 \).
   - David ensures that the set of vertices with degree at least 1 is always in pairs \( (v_{2i-1}, v_{2i}) \). He continues this strategy until all vertices have degree at least 1 and are in disjoint pairs.
   - When all vertices are paired, \( G \) is bipartite with \( |A| = |B| = \frac{n}{2} \). Since \( n \equiv 2 \pmod{4} \), \( \frac{n}{2} \) is odd, making \( \left(\frac{n}{2}\right)^2 \) odd.
   - Before David's turn, the number of edges is even, so \( G \) is not \( K_{|A|, |B|} \). David can always add an edge between \( a \in A \) and \( b \in B \) that has not been drawn, keeping \( G \) bipartite and free of odd cycles.
   - David continues this strategy, ensuring he never creates an odd cycle, and thus he will win.

3. **Case 3: \( 4 \mid n \)**

   - When \( n \) is a multiple of 4, the first player (Jacob) has a winning strategy. Jacob connects two vertices \( v_3 \) and \( v_4 \) not in \( \{v_1, v_2\} \) and employs the same strategy as David in the \( n \equiv 2 \pmod{4} \) case.
   - Jacob ensures that the set of vertices with degree at least 1 is always in pairs \( (v_{2i-1}, v_{2i}) \). He continues this strategy until all vertices have degree at least 1 and are in disjoint pairs.
   - When all vertices are paired, \( G \) is bipartite with \( |A| = |B| = \frac{n}{2} \). Since \( 4 \mid n \), \( \frac{n}{2} \) is even, making \( \left(\frac{n}{2}\right)^2 \) even.
   - Before Jacob's turn, the number of edges is odd, so \( G \) is not \( K_{|A|, |B|} \). Jacob can always add an edge between \( a \in A \) and \( b \in B \) that has not been drawn, keeping \( G \) bipartite and free of odd cycles.
   - Jacob continues this strategy, ensuring he never creates an odd cycle, and thus he will win.

The final answer is \( \boxed{ n } \) is odd or \( 4 \mid n \).