Solution:

The answer is (A). Consider a convex polygon with 2016 sides, and suppose it has exactly $a$ angles of measure less than $150^{\circ}$; the sum of its interior angles equals $(2016 - 2) \cdot 180^{\circ}$. At the same time, since each of the remaining $2016 - a$ angles has measure less than $180^{\circ}$, this sum is strictly less than $a \cdot 150^{\circ} + (2016 - a) \cdot 180^{\circ}$. We thus obtain the inequality
$$2014 \cdot 180^{\circ} < a \cdot 150^{\circ} + (2016 - a) \cdot 180^{\circ},$$
that is, $a \cdot 30^{\circ} < 360^{\circ}$, from which $a < 12$.
On the other hand, it is not difficult to see that there exists a polygon with 11 angles of measure $149^{\circ}$ and 2005 angles of measure $\frac{1}{2005}\left(2014 \cdot 180^{\circ} - 11 \cdot 149^{\circ}\right)$.