Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge1$, find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|$.
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
\[x_k= \left\{ \begin{array}{cl}
{3k}& \text{for $k\le45$,}\\
{-138}& \text{for $k>45$ and $k$ even,}\\
{135}& \text{for $k>45$ and $k$ odd.}
\end{array}
\right.\]