1. Find all integers $n(n \geqslant 2)$, such that for any simple polygon with $2 n$ vertices, there exists a diagonal completely contained within the polygon, which divides the polygon into two polygons with an equal number of vertices.
1. $n=2$.

When $n=2$, it clearly meets the requirement.
When $n \geqslant 3$, we can construct a simple polygon with $2 n$ vertices that does not meet the requirement.

When $n=3$, construct the polygon as shown in Figure 1, which does not meet the requirement.

When $n>3$, add $n-3$ vertices to the segments $B C$ and $E F$ in Figure 1, making them form a nearly straight polyline. The simple polygon thus formed does not meet the requirement.