Given $10$ points in the space such that each $4$ points are not lie on a plane. Connect some points with some segments such that there are no triangles or quadrangles. Find the maximum number of the segments.
To solve the problem of finding the maximum number of segments (edges) that can be drawn between 10 points (vertices) in space such that no three points are collinear and no four points are coplanar, and ensuring that there are no triangles or quadrangles, we can proceed as follows:

1. **Understanding the Problem**:
   - We need to find the maximum number of edges in a graph with 10 vertices.
   - The graph should not contain any 3-cycles (triangles) or 4-cycles (quadrangles).

2. **Graph Theoretical Approach**:
   - We need to construct a graph with 10 vertices and maximize the number of edges while avoiding 3-cycles and 4-cycles.
   - A useful structure to consider is the bipartite graph, which naturally avoids odd cycles (including 3-cycles).

3. **Constructing the Graph**:
   - Consider two sets of 5 vertices each, say \(A = \{a_1, a_2, a_3, a_4, a_5\}\) and \(B = \{b_1, b_2, b_3, b_4, b_5\}\).
   - Connect each vertex in set \(A\) to each vertex in set \(B\). This forms a complete bipartite graph \(K_{5,5}\).

4. **Counting the Edges**:
   - In a complete bipartite graph \(K_{m,n}\), the number of edges is given by \(m \times n\).
   - Here, \(m = 5\) and \(n = 5\), so the number of edges is \(5 \times 5 = 25\).

5. **Verification**:
   - In \(K_{5,5}\), there are no 3-cycles because any cycle must alternate between the two sets \(A\) and \(B\).
   - There are no 4-cycles because any cycle of length 4 would require two vertices from each set, but this would imply a quadrangle, which we are avoiding.

Thus, the maximum number of edges in such a graph is 25.

The final answer is \(\boxed{25}\).