Divide a regular hexagon into $24$ identical small equilateral triangles as shown. To each of the $19$ points that are vertices of at least one of the equilateral triangles, a different number is assigned. Find the maximum possible number of small equilateral triangles with their vertices having numbers in ascending order going clockwise.

[asy]
size(5.5cm);
int n=6;
pair[] V= sequence(new pair(int i){return dir(360*i/n);}, n);
V.cyclic=true;
for(int i=0;i<n;++i){
  draw(V[i]--V[i+1],black+0.7bp);
  draw(V[i]--V[i+3],black+0.7bp);
  draw(midpoint(V[i]--V[i+1])--midpoint(V[i+2]--V[i+3]),black+0.7bp);
}
[/asy]
1. **Define the Problem and Setup:**
   We need to divide a regular hexagon into 24 identical small equilateral triangles. Each of the 19 vertices of these triangles is assigned a unique number. We aim to find the maximum number of small equilateral triangles whose vertices have numbers in ascending order when read clockwise.

2. **Define Good and Bad Triangles:**
   - A triangle is called *good* if its vertices have numbers in ascending order clockwise.
   - A triangle is called *bad* otherwise.

3. **Counting the Number of Violations:**
   Let \( X \) denote the number of times a number in a small triangle is larger than the number immediately clockwise in the same triangle. Note that:
   - In a *bad* triangle, this happens twice.
   - In a *good* triangle, this happens once.

4. **Counting Violations by Edges:**
   Consider the 30 internal edges of the hexagon, which are shared by two small triangles. For each internal edge, regardless of which vertex is larger, there is exactly one small triangle for which moving from the larger to the smaller is a clockwise motion. Thus, the internal edges contribute 30 to \( X \).

5. **Outer Ring Edges:**
   Consider the 12 edges on the outer ring of the hexagon. If none of these edges contributed to \( X \), each vertex in the ring would be less than the number immediately clockwise in the ring, which is impossible. Therefore, at least one of these edges must contribute to \( X \). Thus, the outer ring edges contribute at least 1 to \( X \).

6. **Total Violations:**
   Combining the contributions from the internal and outer edges, we have:
   \[
   X \ge 30 + 1 = 31
   \]

7. **Relating Good and Bad Triangles to Violations:**
   Let \( G \) be the number of good triangles and \( B = 24 - G \) be the number of bad triangles. Each good triangle contributes 1 to \( X \) and each bad triangle contributes 2 to \( X \). Therefore:
   \[
   G + 2B \ge 31
   \]
   Substituting \( B = 24 - G \):
   \[
   G + 2(24 - G) \ge 31
   \]
   Simplifying:
   \[
   G + 48 - 2G \ge 31
   \]
   \[
   -G + 48 \ge 31
   \]
   \[
   -G \ge -17
   \]
   \[
   G \le 17
   \]

8. **Conclusion:**
   The maximum number of good triangles is \( G = 17 \).

The final answer is \(\boxed{17}\).