The answer is 1430. In general, we prove that with $2n$ points the answer is the $n^{\text{th}}$ Catalan number $C_n = \frac{1}{n+1}\binom{2n}{n}$.
First of all, it is well-known that if $S$ is a convex $2n$-gon, then $\chi(S) = C_n$.

It remains to prove the lower bound. We proceed by (strong) induction on $n$, with the base case $n=0$ and $n=1$ clear. Suppose the statement is proven for $0, 1, \dots, n$ and consider a set $S$ with $2(n+1)$ points.
Let $P$ be a point on the convex hull of $S$, and label the other $2n+1$ points $A_1, \dots, A_{2n+1}$ in order of angle from $P$.
Consider drawing a segment $\overline{PA_{2k+1}}$. This splits the $2n$ remaining points into two halves $\mathcal{U}$ and $\mathcal{V}$, with $2k$ and $2(n-k)$ points respectively.

Note that by choice of $P$, no segment in $\mathcal{U}$ can intersect a segment in $\mathcal{V}$. By the inductive hypothesis,
$$\chi(\mathcal{U}) \ge C_k \quad \text{and} \quad \chi(\mathcal{V}) \ge C_{n-k}.$$
Thus, drawing $\overline{PA_{2k+1}}$, we have at least $C_k C_{n-k}$ ways to complete the drawing. Over all choices of $k$, we obtain
$$\chi(S) \ge C_0 C_n + \dots + C_n C_0 = C_{n+1}$$
as desired.