For the convenience of narration, we call two adjacent edges in a graph a "corner".

Lemma: Let $G=(V, E)$ be a simple graph, and graph $G$ is connected, then $G$ contains $\left[\frac{|E|}{2}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$) corners with no common edges.
Proof: We prove by induction on the number of elements in $E$, $|E|$.
When $|E|=0,1,2,3$, the conclusion is obviously true.
Assume $|E| \geqslant 4$, and the conclusion holds for smaller $|E|$. We only need to prove that in graph $G$, we can select two edges $a, b$ to form a corner, and after deleting these two edges $a, b$ from $G$, the remaining graph has a connected component containing $|E|-2$ edges. Applying the induction hypothesis to this connected component yields the conclusion.

Consider the longest path $P: v_{1} v_{2} \cdots v_{k},\left(v_{1}\right.$, $v_{2}, \cdots, v_{k}$ are distinct vertices).
Since graph $G$ is connected, thus, $k \geqslant 3$.
We discuss in cases.
(1) $\operatorname{deg}\left(v_{1}\right) \geqslant 2$.
Since $P$ is the longest path, all neighbors of $v_{1}$ are among $v_{2}, \cdots, v_{k}$, suppose $v_{1} v_{i} \in E(3 \leqslant i \leqslant k)$, then $\left\{v_{1} v_{2}, v_{1} v_{i}\right\}$ is a corner. After deleting these two edges from $E$, if $v_{1}$ has a third edge, the remaining graph is connected; if $v_{1}$ has only the two deleted edges, then $v_{1}$ becomes an isolated vertex, and the rest of the vertices remain connected. In any case, there is a connected component in the remaining graph containing $|E|-2$ edges.
(2) $\operatorname{deg}\left(v_{1}\right)=1, \operatorname{deg}\left(v_{2}\right)=2$.
Then $\left\{v_{1} v_{2}, v_{2} v_{3}\right\}$ is a corner. After deleting these two edges from graph $G$, $v_{1}$ and $v_{2}$ both become isolated vertices, and the rest of the vertices remain connected, thus, there is a connected component containing $|E|-2$ edges.
(3) $\operatorname{deg}\left(v_{1}\right)=1, \operatorname{deg}\left(v_{2}\right) \geqslant 3$, and $v_{2}$ is adjacent to some vertex in $v_{4}, \cdots, v_{k}$.
Then $\left\{v_{1} v_{2}, v_{2} v_{3}\right\}$ is a corner. After deleting these two edges from graph $G$, $v_{1}$ becomes an isolated vertex, and the rest of the vertices remain connected, thus, there is a connected component containing $|E|-2$ edges.
(4) $\operatorname{deg}\left(v_{1}\right)=1, \operatorname{deg}\left(v_{2}\right) \geqslant 3$, and $v_{2}$ is adjacent to some vertex $u \notin\left\{v_{1}, v_{3}, \cdots, v_{k}\right\}$.
Since $P$ is the longest path, thus, all neighbors of $u$ are among $v_{2}, \cdots, v_{k}$. Also, $\left\{v_{1} v_{2}, v_{2} u\right\}$ is a corner. After deleting these two edges from graph $G$, $v_{1}$ becomes an isolated vertex. If $u$ has only the edge $u v_{2}$, then after deleting the said edges, $u$ also becomes an isolated vertex, and the rest of the vertices remain connected; if $u$ has other edges $u v_{i}(3 \leqslant i \leqslant k)$, then after deleting the said edges, all vertices except $v_{1}$ remain connected. In any case, there is a connected component in the remaining graph containing $|E|-2$ edges.
The lemma is proved.
The $V, E$ in the problem can be seen as a graph $G=(V, E)$.
First, we prove: $n \geqslant 2795$.
Let $V=\left\{v_{1}, v_{2}, \cdots, v_{2019}\right\}$.
Among $v_{1}, v_{2}, \cdots, v_{61}$, first connect all pairs of vertices, then delete 15 edges (for example, $v_{1} v_{2}, v_{1} v_{3}, \cdots, v_{1} v_{16}$), a total of $\mathrm{C}_{61}^{2}-15=1815$ edges are connected, thus, the graph formed by these 61 vertices is connected. Then, pair the remaining 2019-61 $=1958$ vertices, and connect one edge between each pair, thus, the graph $G$ has a total of $1815+979=2794$ edges.
From the above construction, we know that any corner in graph $G$ must use the edges connected by $v_{1}, v_{2}, \cdots, v_{61}$. Thus, there are at most $\left[\frac{1815}{2}\right]=907$ corners with no common edges. Therefore, the required $n$ is not less than 2795.
If $|E| \geqslant 2795$, we can arbitrarily delete some edges, and only consider the case $|E|=2795$.
Let graph $G$ have $k$ connected components, each with $m_{1}, m_{2}, \cdots, m_{k}$ vertices, and $e_{1}, e_{2}, \cdots, e_{k}$ edges.
We 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, then $e_{1}, e_{2}, \cdots, e_{k}$ contain at least 981 odd numbers, thus $k \geqslant 981$. Without loss of generality, assume $e_{1}, e_{2}, \cdots, e_{981}$ are all odd, clearly, $m_{1}, m_{2}, \cdots, m_{981} \geqslant 2$.
Let $m=m_{981}+\cdots+m_{k} \geqslant 2$, then
$\mathrm{C}_{m_{i}}^{2} \geqslant e_{i}(1 \leqslant i \leqslant 980), \mathrm{C}_{m}^{2} \geqslant e_{981}+\cdots+e_{k}$,
thus $2795=\sum_{i=1}^{k} e_{i} \leqslant \mathrm{C}_{m}^{2}+\sum_{i=1}^{980} \mathrm{C}_{m_{i}}^{2}$.
Using the convexity of binomial coefficients, i.e., for $x \geqslant y \geqslant 3$,
$$
\mathrm{C}_{x}^{2}+\mathrm{C}_{y}^{2} \leqslant \mathrm{C}_{x+1}^{2}+\mathrm{C}_{y-1}^{2},
$$
we know that when $m_{1}, m_{2}, \cdots, m_{980}, m$ consist of 980 2's and one 59, $\mathrm{C}_{m}^{2}+\sum_{i=1}^{980} \mathrm{C}_{m_{i}}^{2}$ achieves its maximum value. Thus,
$$
\begin{array}{l}
\mathrm{C}_{m}^{2}+\sum_{i=1}^{980} \mathrm{C}_{m_{i}}^{2} \leqslant \mathrm{C}_{59}^{2}+980 \mathrm{C}_{2}^{2} \\
=2691<2795,
\end{array}
$$
which contradicts equation (1). Therefore, $e_{1}, e_{2}, \cdots, e_{k}$ contain at most 979 odd numbers.

Applying the lemma to each connected component, we know that graph $G$ contains $N$ corners with no common edges, where,
$$
\begin{array}{l}
N=\sum_{i=1}^{k}\left[\frac{e_{i}}{2}\right] \geqslant \frac{1}{2}\left(\sum_{i=1}^{k} e_{i}-979\right) \\
=\frac{1}{2}(2795-979)=908 .
\end{array}
$$

In summary, the smallest $n$ is 2795.
(Provided by Zhao Zhiyong)
Four. (50 points) Let $V$ be a set of 2019 points in space, where no four points are coplanar. Some 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 two-element subsets, where each two-element subset consists of two line segments sharing a common endpoint, and any two two-element subsets have an empty intersection.

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