1. Find the minimum number of edges in graph $G$, such that $G$ has $2n$ vertices, and $G$ contains no triangles, and the complement of $G$ has no $n$-vertex complete subgraph.
Let the minimum number of edges be $g(n)$. From the following two figures, we know: $g(4) \leqslant 10, g(n) \leqslant n+5(n \geqslant 5)$.
(1) When $n=2$, $g(n)$ does not exist;
When $n=3$, by the fact that a two-colored graph must contain a triangle, the graph $G$ in the problem does not exist;
When $n=4$, if the number of edges in $G$ $\leqslant 9$, prove that $\bar{G}$ contains $K_{4}$. We consider the cycles in $G$.

First, if $G$ has no odd cycles, then $G$ can be divided into a bipartite graph, where one part must have at least $\frac{2 n}{2}=n=4$ vertices. Taking $n$ vertices, none of which are connected, means $\bar{G}$ contains $K_{4}$. Contradiction! Therefore, $G$ has an odd cycle.
(1) If there is a 7-cycle, let it be $x_{1} \rightarrow x_{2} \rightarrow x_{3} \rightarrow \cdots \rightarrow x_{7}$ $\rightarrow x_{1}$, and the other point is $x_{8}$, then $\bar{G}\left(x_{1}, x_{3}, x_{5}, x_{8}\right), \bar{G}\left(x_{1}\right.$, $\left.x_{3}, x_{6}, x_{8}\right), \bar{G}\left(x_{2}, x_{4}, x_{7}, x_{8}\right)$ must contain at least one $K_{4}$. Note in the above, assume $x_{8} x_{1}, x_{8} x_{3} \notin G(E)$.
(2) If there is a 5-cycle, let it be $x_{1} \rightarrow x_{2} \rightarrow x_{3} \rightarrow x_{4} \rightarrow x_{5}$ $\rightarrow x_{1}$, and the other three points are $x_{6}, x_{7}, x_{8}$. For $\{i, j\} \subset\{6,7,8\}$, if $x_{i}, x_{j}$ are not adjacent, then $x_{i}, x_{j}$ are connected to 3 vertices of the 5-cycle. If $G\left(x_{6}, x_{7}, x_{8}\right)$ has no edges, then $d\left(x_{6}\right)+$ $d\left(x_{7}\right) \geqslant 3, d\left(x_{7}\right)+d\left(x_{8}\right) \geqslant 3, d\left(x_{6}\right)+d\left(x_{8}\right) \geqslant$ 3, so $d\left(x_{6}\right)+d\left(x_{7}\right)+d\left(x_{8}\right) \geqslant 5$, hence the number of edges in $G$ is greater than 9, contradiction!

If $G\left(x_{6}, x_{7}, x_{8}\right)$ has one edge, let $x_{6}, x_{7}$ be adjacent, then $d\left(x_{7}\right)+d\left(x_{8}\right) \geqslant 4 d\left(x_{6}\right)+d\left(x_{8}\right) \geqslant 4, d\left(x_{8}\right)$ $<3$, so the number of edges in $G$ $\geqslant 5+2 \times 4-1-2=10$, contradiction!

If $G\left(x_{6}, x_{7}, x_{8}\right)$ has two edges, let $x_{6} x_{7}, x_{6} x_{8}$ be adjacent, then $d\left(x_{7}\right)+d\left(x_{8}\right) \geqslant 5$, so the number of edges in $G$ $\geqslant 5+$ $5=10$, contradiction! Therefore, $g(4) \geqslant 10$, thus $g(4)=10$.
(2) Proof: In a 10-point graph with 9 edges, there must be 5 points that are not connected to each other, then its complement graph contains $K_{5}$, contradiction!

First, there must be a point with degree $\leqslant 1$, remove this point and its neighbor (if it has no neighbor, then remove any other point), leaving a graph with at most 9 edges and 8 points, so there are 4 points that are not adjacent to each other, plus $V$, we get 5 points that are not adjacent to each other. Hence $g(5) \geqslant$ 10, so $g(5)=10$.
(3) Assume $g(n) \geqslant n+5$, then in a graph with $2(n+1)$ points, following the method in (2), only when $V$ has no neighbors, remove another point with degree not 0, it can be done. Hence $g(n+1) \geqslant n+$ 6. Then $g(n+1)=n+6$.
In summary, $g(4)=10, g(n)=n+5(n \geqslant 5)$.