jmc

The condition $|x_k|=|x_{k-1}+3|$ is equivalent to $x_k^2=(x_{k-1}+3{)}^2$. Thus $$\begin{array}{rl}\sum _{k=1}^{n+1}{x}_k^2& =\sum _{k=1}^{n+1}(x_{k-1}+3{)}^2=\sum _{k=0}^n(x_k+3{)}^2=\left(\sum _{k=0}^n{x}_k^2\right)+\left(6\sum _{k=0}^n{x}_k\right)+9(n+1),\mathrm{so}\\ x_{n+1}^2& =\sum _{k=1}^{n+1}{x}_k^2-\sum _{k=0}^n{x}_k^2=\left(6\sum _{k=0}^n{x}_k\right)+9(n+1),\mathrm{and}\\ \sum _{k=0}^n{x}_k& =\frac{1}{6}\left[x_{n+1}^2-9(n+1)\right].\end{array}$$Therefore, $$\left|\sum _{k=1}^{2006}{x}_k\right|=\frac{1}{6}\left|x_{2007}^2-18063\right|.$$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 $\frac{1}{6}|135^2-18063|=\boxed{27}$, provided there exists a sequence that satisfies the given conditions and for which $x_{2007}=135$.

An example of such a sequence is $$x_k=\{\begin{array}{cl}3k& \text{for k≤45,}\\ -138& \text{for k>45 and k even,}\\ 135& \text{for k>45 and k odd.}\end{array}$$