Japan Mathematical Olympiad

Let $S=x_1+x_2+\cdots +x_{2008}$. 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 $$\begin{array}{rl}{x}_1^2+x_2^2+\cdots +x_{2008}^2& =999^2+(x_1+1{)}^2+\cdots +(x_{2007}+1{)}^2\\ & =(x_1^2+\cdots +x_{2007}^2)+2(x_1+\cdots +x_{2007})+2007+999^2\\ & =(x_1^2+\cdots +x_{2007}^2)+2(S-x_{2008})+1000008,\end{array}$$ 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 $$\begin{gathered} x_n=n-1000(1\le n\le 1000);=-1(1001\le n\le 2008,\text{ and n is odd}); \\ =0(1001\le n\le 2008,\text{ and n is even.}) \end{gathered}$$ 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 $S=x_1+\cdots +x_{2008}$ can take is $-500004$.