Japan Mathematical Olympiad

Let $S = x_1 + x_2 + \dots + x_{2008}$. Since $x_1^2 = |x_1|^2 = 999^2$, and $x_n^2 = |x_{n-1} + 1|^2 = (x_{n-1} + 1)^2$ for $2 \le n \le 2008$, we have
$$\begin{aligned} x_1^2 + x_2^2 + \dots + x_{2008}^2 &= 999^2 + (x_1 + 1)^2 + \dots + (x_{2007} + 1)^2 \\ &= (x_1^2 + \dots + x_{2007}^2) + 2(x_1 + \dots + x_{2007}) + 2007 + 999^2 \\ &= (x_1^2 + \dots + x_{2007}^2) + 2(S - x_{2008}) + 1000008, \end{aligned}$$
from which we obtain
$$2S = x_{2008}^2 + 2x_{2008} - 1000008 = (x_{2008} + 1)^2 - 1000009.$$
Because $|x_n| = |x_{n-1} + 1|$ is satisfied, the even-odd parity of $x_n$ is different from that of $x_{n-1}$, namely, the even-odd parity of $x_n$ changes as $n$ increases by $1$. As $x_1$ is an odd number, $x_{2008}$ is even, and hence, $(x_{2008} + 1)^2 \ge 1$. Consequently, we have $S \ge \frac{1-1000009}{2} = -500004$.
On the other hand, if we set
$$x_n = n - 1000 \quad (1 \le n \le 1000); \quad = -1 \quad (1001 \le n \le 2008, \text{ and $n$ is odd}); \\ = 0 \quad (1001 \le n \le 2008, \text{ and $n$ is even.})$$
then this choice of $x_1, \dots, x_{2008}$ satisfies the conditions of the problem, and since $x_{2008} = 0$ for this choice, we get $S = -500004$. Therefore, the minimum value that $S = x_1 + \dots + x_{2008}$ can take is $-500004$.