11. Given a periodic sequence $\left\{x_{n}\right\}$ satisfying $x_{n}=\left|x_{n-1}-x_{n-2}\right|(n \geqslant 3)$. If $x_{1}=1, x_{2}=a \geqslant 0$. When the smallest positive period of the sequence is minimal, then the sum of the first 2008 terms of the sequence is $\qquad$.
11. Let the sequence $\left\{x_{n}\right\}$ have a period of $T, T \neq 1, T \geqslant 2$. If $T=2$, then $x_{3}=|a-1|=1$, which gives $a=0$ or 2. From $x_{1}=\left|x_{1}-x_{2}\right|=|1-a|=a$, we get $a=\frac{1}{9}$, which contradicts $a=0$ or 2, hence $T \geqslant 3$.
When $T=3$, from $x_{4}=|| a-1|-a|=1$, we get $a \geqslant 1$ or $a=0$. From $x_{5}=|1-| a-1||=a$, we get $0 \leqslant a \leqslant 1$.
Thus, $a=0$ or 1. In this case, the sequence $\left\{x_{n}\right\}$ is either $1,0,1,1,0,1, \cdots$ or $1,1,0,1,1,0, \cdots$, and the sum of any 3 consecutive terms is 2.
Therefore, $S_{2 \times 06}=2 \times 669+1=1339$.
Note: This problem uses a method of verifying each case to prove that the smallest positive period of the sequence $\left\{x_{n}\right\}$ is 3. This step-by-step adjustment method is also a common problem-solving approach in competitions, which helps to cultivate critical and meticulous thinking.