China Southeastern Mathematical Olympiad

Let $x_0 = 2n$, where $n$ is an integer and $|2n| < 1000$, then $|n| \le 499$. So we can choose at most $2 \times 499 + 1 = 999$ numbers, that is, $n \in \{-499, -498, \dots, 0, 1, \dots, 499\}$. Substituting $x_0 = 2n$ into the equation, we get $a = \frac{8n^3 - 1}{2n + 1}$.

Set $f(n) = \frac{8n^3 - 1}{2n + 1}$, then for any $n_1, n_2 \in \{-499, -498, \dots, 0, 1, \dots, 499\}$ ($n_1 \ne n_2$, $n_1, n_2 \in \mathbb{Z}$). If $f(n_1) = f(n_2)$, we can suppose $n_1 = \frac{x_1}{2}$, $n_2 = \frac{x_2}{2}$, where $x_1, x_2$ are roots of $x^3 - a x - a - 1 = 0$. Suppose the other root is $x_3$, then according to the sums and products of roots, we obtain
$$\begin{cases} x_3 = -(x_1 + x_2), \\ x_1 x_2 + x_2 x_3 + x_3 x_1 = -a, \\ x_1 x_2 x_3 = a + 1, \end{cases}$$
that is,
$$\begin{cases} 4N_1 = -a, \\ 8N_2 = a + 1, \end{cases}$$
where $N_1 = -(n_1^2 + n_2^2 + n_1 n_2)$, $N_2 = -n_1 n_2 (n_1 + n_2)$, then $4N_1 + 8N_2 = 1$. Contradiction!
Thus, for any $n_1, n_2 \in \{-499, -498, \dots, 0, 1, \dots, 499\}$, we have $f(n_1) \neq f(n_2)$, which means there are exactly 999 real numbers $a$ that satisfy the condition.

Our aim is to prove that for any even integer $x$ which satisfies $|x| \le 998$, the value of $a = \frac{x^3 - 1}{x + 1}$ is different.
On the contrary, if there exist $x_1 \neq x_2$ that satisfy $\frac{x_1^3 - 1}{x_1 + 1} = \frac{x_2^3 - 1}{x_2 + 1}$, where $x_1, x_2$ are even numbers, then
$$(x_1 - x_2)(x_1^2 x_2 + x_1 x_2^2 + x_2^2 + x_1 x_2 + 1) = 0.$$
Since $x_1 \neq x_2 \Rightarrow x_1 - x_2 \neq 0$ and $x_1^2 x_2 + x_1 x_2^2 + x_1^2 + x_2^2 + x_1 x_2$ is an even number, we obtain
$$(x_1 - x_2)(x_1^2 x_2 + x_1 x_2^2 + x_1^2 + x_2^2 + x_1 x_2 + 1) \neq 0.$$
Contradiction!
Thus, there exist 999 real numbers $a$ that satisfy the condition.