Determine the number of real number $a$, such that for every $a$, equation $x^3=ax+a+1$ has a root $x_0$ satisfying following conditions:
(a) $x_0$ is an even integer;
(b) $|x_0|<1000$.
1. We start with the given equation \( x^3 = ax + a + 1 \). We need to find the number of real numbers \( a \) such that for every \( a \), the equation has a root \( x_0 \) that is an even integer and satisfies \( |x_0| < 1000 \).

2. Let \( x_0 \) be an even integer. Then, we can write the equation as:
   \[
   x_0^3 = ax_0 + a + 1
   \]
   Rearranging this equation to solve for \( a \), we get:
   \[
   a(x_0 + 1) = x_0^3 - 1
   \]
   \[
   a = \frac{x_0^3 - 1}{x_0 + 1}
   \]

3. We need to ensure that \( a \) takes distinct values for different even integers \( x_0 \) within the range \( |x_0| < 1000 \). This means \( x_0 \) can take values from \(-998\) to \(998\) inclusive, but only even integers.

4. To verify that \( a \) takes distinct values, consider two distinct even integers \( x_1 \) and \( x_2 \). Suppose:
   \[
   \frac{x_1^3 - 1}{x_1 + 1} = \frac{x_2^3 - 1}{x_2 + 1}
   \]
   Cross-multiplying, we get:
   \[
   (x_1^3 - 1)(x_2 + 1) = (x_2^3 - 1)(x_1 + 1)
   \]
   Expanding both sides:
   \[
   x_1^3 x_2 + x_1^3 - x_2 - 1 = x_2^3 x_1 + x_2^3 - x_1 - 1
   \]
   Rearranging terms:
   \[
   x_1^3 x_2 + x_1^3 - x_2 = x_2^3 x_1 + x_2^3 - x_1
   \]
   \[
   x_1^3 x_2 - x_2^3 x_1 = x_1 - x_2
   \]
   Factoring out common terms:
   \[
   x_2(x_1^3 - x_2^2 x_1) = x_1(x_2^3 - x_1^2 x_2)
   \]
   \[
   x_2 x_1 (x_1^2 - x_2^2) = x_1 x_2 (x_2^2 - x_1^2)
   \]
   Since \( x_1 \neq x_2 \), this implies:
   \[
   x_1^2 - x_2^2 = x_2^2 - x_1^2
   \]
   Which is a contradiction unless \( x_1 = x_2 \). Therefore, \( a \) must take distinct values for different even integers \( x_0 \).

5. The number of even integers \( x_0 \) within the range \( |x_0| < 1000 \) is from \(-998\) to \(998\), inclusive. This gives us:
   \[
   \text{Number of even integers} = \frac{1998}{2} + 1 = 999
   \]

The final answer is \(\boxed{999}\).