4. Let $V$ be a set of 2019 points in space, no four of which are coplanar. Some of these points are connected by line segments, and let $E$ be the set of these line segments. Find the smallest positive integer $n$ such that if $E$ has at least $n$ elements, then $E$ must contain 908 pairwise disjoint subsets, each consisting of two line segments that share a common endpoint.
To facilitate the narrative, we call two adjacent edges in a graph a "corner." First, we prove a lemma: Let \( G = (V, E) \) be a simple graph, and \( G \) is connected, then \( G \) contains \(\left[\frac{|E|}{2}\right]\) corners with no common edges (here \([\alpha]\) denotes the integer part of the real number \(\alpha\)).

We prove this by induction on the number of elements in \( E \), \( |E| \). When \( |E| = 0, 1, 2, 3 \), the conclusion is obviously true. Assume \( |E| \geq 4 \), and the conclusion holds for smaller \( |E| \). We need to show that in \( G \), we can select two edges \( a, b \) to form a corner, and after deleting \( a, b \) from \( G \), the remaining graph has a connected component with \( |E| - 2 \) edges. Applying the induction hypothesis to this connected component will complete the proof.

Consider the longest path \( P: v_1 v_2 \cdots v_k \) in \( G \), where \( v_1, v_2, \cdots, v_k \) are distinct vertices. Since \( G \) is connected, \( k \geq 3 \).

Case 1: \(\operatorname{deg}(v_1) \geq 2\). Since \( P \) is the longest path, all neighbors of \( v_1 \) are among \( v_2, v_3, \cdots, v_k \). Let \( v_1 v_i \in E \) where \( 3 \leq i \leq k \), then \(\{v_1 v_2, v_1 v_i\}\) is a corner. After deleting these two edges, if \( v_1 \) has a third edge, the remaining graph is connected; if \( v_1 \) has only the two deleted edges, \( v_1 \) becomes an isolated vertex, and the rest of the vertices remain connected. In either case, the remaining graph has a connected component with \( |E| - 2 \) edges.

Case 2: \(\operatorname{deg}(v_1) = 1, \operatorname{deg}(v_2) = 2\). Then \(\{v_1 v_2, v_2 v_3\}\) is a corner. After deleting these two edges, \( v_1 \) and \( v_2 \) become isolated vertices, and the rest of the vertices remain connected, so there is a connected component with \( |E| - 2 \) edges.

Case 3: \(\operatorname{deg}(v_1) = 1, \operatorname{deg}(v_2) \geq 3\), and \( v_2 \) is adjacent to some vertex in \( v_4, \cdots, v_k \). Then \(\{v_1 v_2, v_2 v_3\}\) is a corner. After deleting these two edges, \( v_1 \) becomes an isolated vertex, and the rest of the vertices remain connected, so there is a connected component with \( |E| - 2 \) edges.

Case 4: \(\operatorname{deg}(v_1) = 1, \operatorname{deg}(v_2) \geq 3\), and \( v_2 \) is adjacent to some vertex \( u \notin \{v_1, v_2, \cdots, v_k\} \). Since \( P \) is the longest path, all neighbors of \( u \) are among \( v_2, v_3, \cdots, v_k \). Since \(\{v_1 v_2, v_2 u\}\) is a corner, after deleting these two edges, \( v_1 \) becomes an isolated vertex. If \( u \) has only the edge \( u v_2 \), then after deleting the edge, \( u \) also becomes an isolated vertex, and the rest of the vertices remain connected; if \( u \) has other edges \( u v_i, 3 \leq i \leq k \), then after deleting the edge, the rest of the vertices remain connected. In either case, the remaining graph has a connected component with \( |E| - 2 \) edges. The lemma is proved.

Returning to the original problem, consider \( V \) and \( E \) as a graph \( G = (V, E) \). First, we prove: \( n \geq 2795 \). Let \( V = \{v_1, v_2, \cdots, v_{2019}\} \). In \( v_1, v_2, \cdots, v_{61} \), first connect all pairs of vertices, then delete 15 edges (e.g., \( v_1 v_2, v_1 v_3, \cdots, v_1 v_{16} \)), resulting in \( C_{61}^2 - 15 = 1815 \) edges, forming a connected graph. Then, pair the remaining \( 2019 - 61 = 1958 \) vertices, connecting each pair with one edge, resulting in \( 1815 + 979 = 2794 \) edges in graph \( G \).

From the above construction, any corner in \( G \) must use edges connected among \( v_1, v_2, \cdots, v_{61} \), so there are at most \(\left[\frac{1815}{2}\right] = 907\) corners with no common edges. Therefore, \( n \) must be at least 2795.

On the other hand, if \( |E| \geq 2795 \), we can arbitrarily delete some edges, so we only need to consider the case \( |E| = 2795 \). Suppose \( G \) has \( k \) connected components, with \( m_1, m_2, \cdots, m_k \) vertices and \( e_1, e_2, \cdots, e_k \) edges, respectively. We need to prove: \( e_1, e_2, \cdots, e_k \) contain at most 979 odd numbers.

Proof by contradiction. Assume \( e_1, e_2, \cdots, e_k \) contain at least 980 odd numbers. Since \( e_1 + e_2 + \cdots + e_k = 2795 \) is odd, \( e_1, e_2, \cdots, e_k \) must contain at least 981 odd numbers, so \( k \geq 981 \).

Assume \( e_1, e_2, \cdots, e_{981} \) are all odd, and clearly \( m_1, m_2, \cdots, m_{981} \geq 2 \).

Let \( m = m_{981} + \cdots + m_k \geq 2 \), then \( C_{m_i}^2 \geq e_i (1 \leq i \leq 980) \), \( C_m^2 > e_{981} + \cdots + e_k \), so \( 2795 = \sum_{i=1}^k e_i \leq C_m^2 + \sum_{i=1}^{980} C_{m_i}^2 \).

Using the convexity of binomial coefficients, i.e., for \( x \geq y \geq 3 \), \( C_x^2 + C_y^2 \leq C_{x+1}^2 + C_{y-1}^2 \), the maximum value of \( C_m^2 + \sum_{i=1}^{980} C_{m_i}^2 \) is achieved when \( m_1, m_2, \cdots, m_{980}, m \) consist of 980 twos and one 59.

Thus, \( C_m^2 + \sum_{i=1}^{980} C_{m_i}^2 \leq C_{59}^2 + 980 C_2^2 = 2691 < 2795 \), a contradiction.

Therefore, \( e_1, e_2, \cdots, e_k \) contain at most 979 odd numbers.

Applying the lemma to each connected component, \( G \) contains \( N \) corners with no common edges, where \( N = \sum_{i=1}^k \left[\frac{e_i}{2}\right] \geq \frac{1}{2} \left( \sum_{i=1}^k e_i - 979 \right) = \frac{1}{2} (2795 - 979) = 908 \).

In conclusion, the minimum value of \( n \) is 2795.