Let $\mathcal S$ be a set of $16$ points in the plane, no three collinear. Let $\chi(S)$ denote the number of ways to draw $8$ lines with endpoints in $\mathcal S$, such that no two drawn segments intersect, even at endpoints. Find the smallest possible value of $\chi(\mathcal S)$ across all such $\mathcal S$. 

[i]Ankan Bhattacharya[/i]
1. **Define the Problem and Variables:**
   Let $\mathcal{S}$ be a set of 16 points in the plane, with no three points collinear. We need to find the smallest possible value of $\chi(\mathcal{S})$, where $\chi(\mathcal{S})$ denotes the number of ways to draw 8 non-intersecting line segments with endpoints in $\mathcal{S}$.

2. **Catalan Numbers:**
   The problem can be related to the Catalan numbers, which count the number of ways to draw non-intersecting chords between points on a circle. The $n$-th Catalan number $C_n$ is given by:
   \[
   C_n = \frac{1}{n+1} \binom{2n}{n}
   \]
   For $n = 8$, we need to find $C_8$.

3. **Recursive Formula for Catalan Numbers:**
   The Catalan numbers can also be defined recursively as:
   \[
   C_n = \sum_{k=0}^{n-1} C_k C_{n-1-k}
   \]
   This recursive definition helps in understanding the structure of non-intersecting segments.

4. **Convex Hull and Non-Intersecting Segments:**
   If the points in $\mathcal{S}$ form a convex $2n$-gon, the number of ways to draw $n$ non-intersecting segments is exactly the $n$-th Catalan number. This is because any segment drawn inside a convex polygon will not intersect with another segment if drawn correctly.

5. **Calculation of $C_8$:**
   Using the formula for Catalan numbers:
   \[
   C_8 = \frac{1}{8+1} \binom{16}{8} = \frac{1}{9} \binom{16}{8}
   \]
   Calculate $\binom{16}{8}$:
   \[
   \binom{16}{8} = \frac{16!}{8!8!} = 12870
   \]
   Therefore:
   \[
   C_8 = \frac{12870}{9} = 1430
   \]

6. **Conclusion:**
   The smallest possible value of $\chi(\mathcal{S})$ across all such $\mathcal{S}$ is $C_8 = 1430$.

The final answer is $\boxed{1430}$.