All vertices of a regular 2016-gon are initially white. What is the least number of them that can be painted black so that:\\
(a) There is no right triangle\\
(b) There is no acute triangle\\
having all vertices in the vertices of the 2016-gon that are still white?
To solve this problem, we need to determine the minimum number of vertices that must be painted black in a regular 2016-gon so that no right or acute triangles can be formed with the remaining white vertices.

### Part (b): No Acute Triangle

1. **Understanding the Problem:**
   - We need to ensure that no three white vertices form an acute triangle.
   - An acute triangle has all its angles less than \(90^\circ\).

2. **Key Insight:**
   - In a regular polygon, the largest possible chord is the diameter.
   - If we consider the vertices along a diameter, any triangle formed by these vertices will not be acute.

3. **Strategy:**
   - We can leave at most all the vertices over one diagonal plus one of the points in the diagonal.
   - This is because if we leave any vertices outside this configuration, we can form an acute triangle.

4. **Proof:**
   - Consider the greatest chord \(PQ\) that we can have with the points as ends.
   - Let \(P'\) and \(Q'\) be their symmetric points with respect to the circumcenter.
   - If there exists any point in the minor arcs \(PA\) and \(P'Q'\), then we will be able to form a chord greater than \(PQ\), which is impossible.
   - Moreover, if we have any vertices on the circle on the opposite side of \(P\) with respect to \(P'Q'\), then we can form an acute triangle.
   - Therefore, all the points left white must be over the diagonal.

5. **Calculation:**
   - A regular 2016-gon has 2016 vertices.
   - The diagonal divides the polygon into two equal halves, each containing 1008 vertices.
   - To ensure no acute triangles, we can leave at most 1008 vertices white.

Thus, the least number of vertices that must be painted black is \(2016 - 1008 = 1008\).

The final answer is \(\boxed{1008}\).