Is this even possible?

Here is the general solution for this problem (here, $n=9$ ):

Let the number of ways for $2n$ points be $a_{n}$ . Choose one of the points and label it $M$ . Then, $M$ must be joined to a point, say $N$ , such that there are even number of points on either side of the segment $MN$ . This gives us the recurrence relation (yeah, another recurrence XD) : $a_{n} = a_{0}a_{n-1}+a_{1}a_{n-2}+...+a_{n-1}a_{0}, a_{0}=1$ Let $b_{n} = a_{n-1}$ where $n\geq 1$ . Thus, $b_{n} = b_{1}b_{n-1}+b_{2}b_{n-2}+...+b_{n-1}b_{1}, b_{1}=1$ Hence, we obtain the equations $b_{n}=\dfrac{1}{n}\dbinom{2n-2}{n-1}$ and $a_{n}=b_{n+1}$ which is the same as the $n^\text{th}$ Catalan number.

We assign values to the points in a order like this $1,2,3,.....18$ . Now, suppose that whenever two points $1\leq i<j\leq18$ are connected with a line segment we say that we give $i$ the value $1$ and $j$ the value $-1$ . And, its notable that if no two such line segments $l_{ij}$ doesn't intersect there will be a the property like this $a_{1}+a_{2}+....a_{k}\geq 0$ fro all $k \leq 18$ . Hence, there are total $C_9=4862$ ways of getting this.