I think the answer is
1
Proof
Let us consider the island as a unit circle in the complex plane such that the nine posts are located along the circumference and are represented by complex numbers of unit length. Let us consider an arbitrary grouping of these nine complex numbers into three groups which will define the vertices of our three triangles - $(A_1, A_2, A_3), \,\,(B_1, B_2, B_3),\,\, (C_1, C_2, C_3)$
Then, as the circumcentre of each triangle is located at the origin, the orthocentres of these triangles in the complex plane are just given by the sum of the vertices, i.e, $$ O_1 = A_1 + A_2 + A_3 \,\,,\,\, O_2 = B_1 + B_2 + B_3 \,\,,\,\, O_3 = C_1 + C_2 + C_3$$
Then, the centroid formed by these three points is just the average of the coordinates, that is $$ C = \frac{O_1 + O_2 + O_3}{3} = \frac{A_1 + A_2 + A_3 + B_1 + B_2 + B_3 + C_1 + C_2 + C_3}{3}$$ Notice, now that this is independent of our choice of triangles and is located at the point (considering $i$ as being the Northern post) $$C = \frac{2 \cos\left( \frac{7\pi}{8}\right) + \cos\left(\frac{\pi}{4} \right) + i \sin \left( \frac{\pi}{4}\right)}{3}$$ in the complex plane.
It just remains to show that there is a choice of triangles such that the orthocentres are on the island. We can use the following choice, now using directions $$(N,S,W),\,\, (NNE, SSW, WNW),\,\, (NE, WSW, E)$$ The orthocentres of the first two triangles are at the points $W$ and $WNW$, respectively. To find the orthocentre of the third triangle, we resort to the complex plane representation once more to find that the orthocentre is located at $$ O_3 = \left(1-\cos\left(\frac{\pi}{8} \right) + \cos \left(\frac{\pi}{4}\right)\right) + i \left( \sin\left( \frac{\pi}{4}\right) - \sin \left( \frac{\pi}{8}\right)\right) \approx 0.783 + 0.324i$$ which is on the island since it has magnitude less than $1$.
