Answer: 27

Solution:

Squaring the recurrence gives $x_k^2 = (x_{k-1} + 3)^2$ $= x_{k-1}^2 + 6 x_{k-1} + 9.$ Summing for $k = 1$ to $2007$ telescopes: $$x_{2007}^2 = x_0^2 + 6 \sum_{k=0}^{2006} x_k + 9 \cdot 2007,$$ so with $x_0 = 0,$ $$\begin{aligned} &x_1 + x_2 + \cdots + x_{2006} \\ &= \frac{x_{2007}^2 - 18063}{6}. \end{aligned}$$

Induction shows each $x_k$ is a multiple of $3$ whose parity matches that of $k,$ so $x_{2007}$ is an odd multiple of $3.$ To minimize $\left|x_{2007}^2 - 18063\right|,$ take the odd multiple of $3$ whose square is nearest $18063:$ that is $\pm 135,$ with $135^2 = 18225,$ giving $\frac{18225 - 18063}{6} = 27$ (the neighbors $129$ and $141$ give $237$ and $303$).

This value is attained: take $x_k = 3k$ for $k \le 45,$ and thereafter alternate $x_k = -138$ for even $k$ and $x_k = 135$ for odd $k;$ then $x_{2007} = 135$ and the sum is $27.$ So the minimum is $27.$