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 $x^3-ax-a-1=0$. Suppose the other root is $x_3$, then according to the sums and products of roots, we obtain $$\{\begin{array}{l}{x}_3=-(x_1+x_2),\\ x_1{x}_2+x_2{x}_3+x_3{x}_1=-a,\\ x_1{x}_2{x}_3=a+1,\end{array}$$ that is, $$\{\begin{array}{l}4N_1=-a,\\ 8N_2=a+1,\end{array}$$ 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 $f(n_1)\ne f(n_2)$, which means there are exactly 999 real numbers $a$ that satisfy the condition.

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Alternative solution.

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 $x_1\ne x_2$ 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 $x_1\ne x_2\Rightarrow x_1-x_2\ne 0$ 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 $$(x_1-x_2)(x_1^2{x}_2+x_1{x}_2^2+x_1^2+x_2^2+x_1{x}_2+1)\ne 0.$$ Contradiction! Thus, there exist 999 real numbers $a$ that satisfy the condition.