The condition $ |x_k|=|x_{k-1}+3| $ is equivalent to $ x_k^2=(x_{k-1}+3)^2 $. Thus $$\begin{aligned}\sum_{k=1}^{n+1}x_k^2&=\sum_{k=1}^{n+1}(x_{k-1}+3)^2 =\sum_{k=0}^{n}(x_{k}+3)^2 =\left(\sum_{k=0}^{n}x_k^2\right) +\left(6\sum_{k=0}^{n}x_k\right)+9(n+1),\quad{\rm so}\cr x_{n+1}^2&=\sum_{k=1}^{n+1}x_k^2 -\sum_{k=0}^{n}x_k^2 =\left(6\sum_{k=0}^{n}x_k\right)+9(n+1),\quad{\rm and}\cr \sum_{k=0}^{n}x_k&= {1\over6}\left[x_{n+1}^2-9(n+1)\right].\end{aligned}$$Therefore, \[\displaystyle \left|\sum_{k=1}^{2006}x_k\right| ={1\over6}\left|x_{2007}^2-18063\right|.\]Notice that $ x_k $ is a multiple of 3 for all $ k $, and that $ x_k $ and $ k $ have the same parity. The requested sum will be a minimum when $ |x_{2007}^2-18063| $ is a minimum, that is, when $ x_{2007} $ is the multiple of 3 whose square is as close as possible to 18063. Check odd multiples of 3, and find that $ 129^2<16900 $, $ 141^2>19600 $, and $ 135^2=18225 $. The requested minimum is therefore $ {1\over6}|135^2-18063|=\boxed{27} $, provided there exists a sequence that satisfies the given conditions and for which $ x_{2007}=135 $. An example of such a sequence is \[x_k= \left\{ \begin{array}{cl} {3k}& \text{for $ k\le45 $,}\\ {-138}& \text{for $ k>45 $ and $ k $ even,}\\ {135}& \text{for $ k>45 $ and $ k $ odd} \end{array} \right.\]