A continuous function $f : R  \to R$ satisfies the conditions $f(1000) = 999$ and $f(x)f(f(x)) = 1$ for all real $x$. Determine $f(500)$.
1. Given the function \( f : \mathbb{R} \to \mathbb{R} \) is continuous and satisfies the conditions \( f(1000) = 999 \) and \( f(x)f(f(x)) = 1 \) for all real \( x \).

2. From the condition \( f(x)f(f(x)) = 1 \), we can derive:
   \[
   f(f(x)) = \frac{1}{f(x)}
   \]

3. We need to determine \( f(500) \). To do this, we will show that there exists some \( t \in \mathbb{R} \) such that \( f(t) = 500 \).

4. Assume for contradiction that \( f(x) < 500 \) for all \( x \in \mathbb{R} \). This would imply that \( f(1000) = 999 \) is a contradiction because \( 999 > 500 \). Therefore, \( f(x) < 500 \) for all \( x \in \mathbb{R} \) is not possible.

5. Similarly, assume for contradiction that \( f(x) > 500 \) for all \( x \in \mathbb{R} \). Then, from \( f(f(x)) = \frac{1}{f(x)} \), we would have:
   \[
   f(f(x)) < \frac{1}{500}
   \]
   This would imply that \( f(f(x)) \) is always less than \( \frac{1}{500} \). However, since \( f \) is continuous and \( f(1000) = 999 \), this would again lead to a contradiction because \( f(f(1000)) = f(999) \) should be greater than \( \frac{1}{500} \).

6. Since both assumptions lead to contradictions, there must exist some \( t \in \mathbb{R} \) such that \( f(t) = 500 \).

7. Let \( x = t \). Then:
   \[
   f(500) = \frac{1}{500}
   \]

The final answer is \( \boxed{\frac{1}{500}} \).