There are $m \ge 2$ blue points and $n \ge 2$ red points in three-dimensional space, and no four points are coplanar. Geoff and Nazar take turns, picking one blue point and one red point and connecting the two with a straight-line segment. Assume that Geoff starts first and the one who first makes a cycle wins. Who has the winning strategy? 
1. **Translation to Graph Theory**:
   - We can translate the problem into a graph theory problem where blue points and red points are vertices of two different sets in a bipartite graph. Geoff and Nazar take turns adding edges between these two sets.
   - The goal is to determine who has a winning strategy, i.e., who can force the creation of a cycle first.

2. **Winning Strategy for Geoff**:
   - Geoff wins if both \( m \) and \( n \) are odd.
   - Geoff starts by drawing any edge. Let's analyze Nazar's possible moves:
     - If Nazar draws an edge that completes a path of length 3, Geoff immediately wins by completing the cycle.
     - If Nazar draws an edge \((b, r)\) where \( b \) is dirty (already has an edge) and \( r \) is clean, Geoff can draw an edge \((b, r')\) where \( r' \) is another clean vertex of the same color as \( r \).
     - If Nazar draws an edge \((b, r)\) where both vertices are clean, Geoff can draw another edge \((b', r')\) where both vertices are also clean.
   - The key observation is that after Geoff's first move, there are an even number of clean vertices of each color remaining. Whenever Nazar uses a clean vertex, it leaves an odd number of clean vertices of that color, ensuring Geoff can always find another clean vertex to use.
   - Geoff will not be the first to create a path of length 3, forcing Nazar to do so and thus giving Geoff the win.

3. **Winning Strategy for Nazar**:
   - If at least one of \( m \) or \( n \) is even, Nazar wins. Without loss of generality, assume \( n \) (the number of red points) is even.
   - Nazar's strategy is to ensure that all dirty blue vertices have a degree of at least 2. If Geoff makes a move \((b, r)\) where \( b \) is blue and \( r \) is red, Nazar will:
     - Either win immediately if Geoff's move creates a path of length 3.
     - Or play \((b, r')\) where \( r' \) is a fresh red vertex.
   - This strategy ensures that all dirty blue vertices have edges connecting to at least two red vertices. If Geoff uses one of these red vertices, it creates a path of length 3, giving Nazar the win.
   - Since there are an even number of red vertices, and each move by Geoff is matched by a move from Nazar, Nazar will always have a clean red vertex available, ensuring he can continue his strategy and eventually win.

The final answer is Geoff wins if \( \boxed{ m } \) and \( n \) are both odd, and Nazar wins otherwise.