How many triangles formed by three vertices of a regular $17$-gon are obtuse?

$\text{(A) }156\qquad\text{(B) }204\qquad\text{(C) }357\qquad\text{(D) }476\qquad\text{(E) }524$
1. **Lemma:** A triangle formed by three vertices of a regular polygon is obtuse if and only if all three points are on the same half of the circumcircle.

   **Proof:** Inscribe the polygon in its circumcircle. The largest angle in a triangle will be obtuse if and only if the arc it subtends is greater than \(180^\circ\). This means that all three vertices of the triangle must lie on the same semicircle of the circumcircle.

2. **Choosing the vertices:**
   - Consider a regular 17-gon inscribed in a circle.
   - Label the vertices \(A_1, A_2, \ldots, A_{17}\) in a clockwise manner.

3. **Fixing a vertex:**
   - Fix one vertex, say \(A_1\). 
   - Draw the diameter through \(A_1\). This diameter will divide the circle into two semicircles.
   - The vertices on the same semicircle as \(A_1\) are \(A_2, A_3, \ldots, A_9\) (8 vertices).

4. **Choosing the other two vertices:**
   - To form an obtuse triangle with \(A_1\), the other two vertices must be chosen from the 8 vertices on the same semicircle.
   - The number of ways to choose 2 vertices from these 8 vertices is given by the binomial coefficient \(\binom{8}{2}\).

5. **Calculating the number of obtuse triangles:**
   - The number of ways to choose 2 vertices from 8 is:
     \[
     \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28
     \]
   - Since there are 17 vertices in total, and we can fix any one of them as \(A_1\), the total number of obtuse triangles is:
     \[
     17 \times 28 = 476
     \]

Conclusion:
\[
\boxed{476}
\]