The degree measures of the six interior angles of a convex hexagon form an arithmetic sequence (not necessarily in cyclic order).  The common difference of this arithmetic sequence can be any real number in the open interval $(-D, D)$.  Compute the greatest possible value of $D$.
1. Let \( d \) be the common difference of the arithmetic sequence of the interior angles of the convex hexagon. Assume without loss of generality (WLOG) that \( d \ge 0 \) (if \( d < 0 \), we can consider the absolute value of \( d \)).

2. The sum of the interior angles of a convex hexagon is given by:
   \[
   180^\circ \times (6 - 2) = 720^\circ
   \]
   Since there are six angles, the average measure of each angle is:
   \[
   \frac{720^\circ}{6} = 120^\circ
   \]

3. The angles form an arithmetic sequence, so we can denote the angles as:
   \[
   120^\circ - \frac{5d}{2}, 120^\circ - \frac{3d}{2}, 120^\circ - \frac{d}{2}, 120^\circ + \frac{d}{2}, 120^\circ + \frac{3d}{2}, 120^\circ + \frac{5d}{2}
   \]

4. For the hexagon to be convex, all interior angles must be between \( 0^\circ \) and \( 180^\circ \). Therefore, the smallest angle must be greater than \( 0^\circ \) and the largest angle must be less than \( 180^\circ \).

5. The smallest angle is:
   \[
   120^\circ - \frac{5d}{2} > 0^\circ
   \]
   Solving for \( d \):
   \[
   120^\circ > \frac{5d}{2}
   \]
   \[
   240^\circ > 5d
   \]
   \[
   48^\circ > d
   \]
   \[
   d < 48^\circ
   \]

6. The largest angle is:
   \[
   120^\circ + \frac{5d}{2} < 180^\circ
   \]
   Solving for \( d \):
   \[
   120^\circ + \frac{5d}{2} < 180^\circ
   \]
   \[
   \frac{5d}{2} < 60^\circ
   \]
   \[
   5d < 120^\circ
   \]
   \[
   d < 24^\circ
   \]

7. Combining both conditions, we get:
   \[
   d < 24^\circ
   \]

8. Therefore, the greatest possible value of \( D \) is:
   \[
   \boxed{24}
   \]