Let $S$ be a set of $2017$ points in the plane. Let $R$ be the radius of the smallest circle containing all points in $S$ on either the interior or boundary. Also, let $D$ be the longest distance between two of the points in $S$. Let $a$, $b$ be real numbers such that $a\le \frac{D}{R}\le b$ for all possible sets $S$, where $a$ is as large as possible and $b$ is as small as possible. Find the pair $(a, b)$.
1. **Determine the value of \( b \):**
   - The value of \( b \) is determined by the scenario where the longest distance \( D \) is maximized relative to the radius \( R \) of the smallest circle containing all points.
   - Consider two points \( A \) and \( B \) in the set \( S \). The smallest circle containing these two points has them as diametrically opposite points.
   - Therefore, the diameter of the circle is \( D \), and the radius \( R \) is \( \frac{D}{2} \).
   - Hence, \( \frac{D}{R} = \frac{D}{\frac{D}{2}} = 2 \).
   - This shows that \( b = 2 \).

2. **Determine the value of \( a \):**
   - To find the minimum value of \( \frac{D}{R} \), consider a set of points forming an equilateral triangle.
   - Let the side length of the equilateral triangle be \( s \). The longest distance \( D \) between any two points is \( s \).
   - The radius \( R \) of the circumscribed circle around the equilateral triangle is \( \frac{s}{\sqrt{3}} \).
   - Therefore, \( \frac{D}{R} = \frac{s}{\frac{s}{\sqrt{3}}} = \sqrt{3} \).
   - This shows that \( a = \sqrt{3} \).

3. **Verify the bounds:**
   - For any set of points, the value of \( \frac{D}{R} \) must lie between \( \sqrt{3} \) and \( 2 \).
   - The value \( \sqrt{3} \) is achieved with an equilateral triangle, and the value \( 2 \) is achieved with two points.

Thus, the pair \( (a, b) \) is \( (\sqrt{3}, 2) \).

The final answer is \( \boxed{ (\sqrt{3}, 2) } \).