Let $S$ be the set of all real numbers $x$ such that $0 \le x \le 2016 \pi$ and $\sin x < 3 \sin(x/3)$.  The set $S$ is the union of a finite number of disjoint intervals.  Compute the total length of all these intervals.
1. We start with the given inequality $\sin x < 3 \sin(x/3)$. To analyze this, we use the triple angle identity for sine:
   \[
   \sin x = 3 \sin(x/3) - 4 \sin^3(x/3)
   \]
   Therefore, the inequality becomes:
   \[
   3 \sin(x/3) - 4 \sin^3(x/3) < 3 \sin(x/3)
   \]

2. Simplifying the inequality, we subtract $3 \sin(x/3)$ from both sides:
   \[
   3 \sin(x/3) - 4 \sin^3(x/3) - 3 \sin(x/3) < 0
   \]
   \[
   -4 \sin^3(x/3) < 0
   \]

3. Dividing both sides by $-4$ (which reverses the inequality):
   \[
   \sin^3(x/3) > 0
   \]
   Since $\sin^3(x/3) > 0$, it implies:
   \[
   \sin(x/3) > 0
   \]

4. The function $\sin(x/3)$ is positive in the intervals where $x/3$ is in the first and second quadrants of the unit circle, i.e., $0 < x/3 < \pi$ and $2\pi < x/3 < 3\pi$, and so on. Therefore, $x$ must be in the intervals:
   \[
   0 < x < 3\pi, \quad 6\pi < x < 9\pi, \quad 12\pi < x < 15\pi, \quad \ldots
   \]

5. Each interval where $\sin(x/3) > 0$ has a length of $3\pi$. Since $x$ ranges from $0$ to $2016\pi$, we need to determine how many such intervals fit within this range. The period of $\sin(x/3)$ is $6\pi$, so we divide $2016\pi$ by $6\pi$ to find the number of periods:
   \[
   \frac{2016\pi}{6\pi} = 336
   \]

6. Each period of $6\pi$ contains one interval of length $3\pi$ where $\sin(x/3) > 0$. Therefore, the total length of all these intervals is:
   \[
   336 \times 3\pi = 1008\pi
   \]

The final answer is $\boxed{1008\pi}$