4. Given $P_{1}, P_{2}, \cdots, P_{100}$ as 100 points on a plane, satisfying that no three points are collinear. For any three of these points, if their indices are in increasing order and they form a clockwise orientation, then the triangle with these three points as vertices is called "clockwise". Question: Is it possible to have exactly
$$
2017 \text { clockwise triangles? }
$$
4. Suppose $P_{1}, P_{2}, \cdots, P_{100}$ are arranged counterclockwise on a circle. At this point, the number of clockwise triangles is 0.

Now, move these points (not necessarily along the circumference). When a point $P_{i}$ crosses the line $P_{j} P_{k}$ during the movement, the orientation (clockwise or counterclockwise) of $\triangle P_{i} P_{j} P_{k}$ will change. It is stipulated that during the movement, only one point is allowed to cross the line connecting two other points, and this operation does not change the orientation of any other triangle. Continuously perform such operations until $P_{1}, P_{2}, \cdots, P_{100}$ are arranged clockwise on the circle. At this point, the number of clockwise triangles is $\mathrm{C}_{100}^{3}>2017$.

Notice that, after each operation, the number of clockwise triangles increases by 1. Therefore, there must be a moment when the number of clockwise triangles is exactly 2017.