Given three distinct positive integers such that one of them is the average of the two others. Can the product of these three integers be the perfect 2008th power of a positive integer?
1. **Define the integers:**
   Let the three distinct positive integers be \( x \), \( y \), and \( z \). Without loss of generality, assume \( y \) is the average of \( x \) and \( z \). Therefore, we have:
   \[
   y = \frac{x + z}{2}
   \]
   Since \( x \), \( y \), and \( z \) are integers, \( x + z \) must be even, implying \( x \) and \( z \) must have the same parity (both even or both odd).

2. **Express the product:**
   We need to determine if the product \( x \cdot y \cdot z \) can be a perfect 2008th power of a positive integer. Let:
   \[
   P = x \cdot y \cdot z
   \]
   We need \( P \) to be of the form \( k^{2008} \) for some integer \( k \).

3. **Choose specific forms for \( x \), \( y \), and \( z \):**
   Consider the given example:
   \[
   x = a^{669}(a-b)^{670}(a+b)^{669}
   \]
   \[
   y = a^{670}(a-b)^{669}(a+b)^{669}
   \]
   \[
   z = a^{669}(a-b)^{669}(a+b)^{670}
   \]
   where \( a \) and \( b \) are positive integers with \( a > b \).

4. **Verify the average condition:**
   Check if \( y \) is indeed the average of \( x \) and \( z \):
   \[
   y = \frac{x + z}{2}
   \]
   Substitute the expressions for \( x \), \( y \), and \( z \):
   \[
   a^{670}(a-b)^{669}(a+b)^{669} = \frac{a^{669}(a-b)^{670}(a+b)^{669} + a^{669}(a-b)^{669}(a+b)^{670}}{2}
   \]
   Simplify the right-hand side:
   \[
   a^{670}(a-b)^{669}(a+b)^{669} = \frac{a^{669}(a-b)^{669}[(a-b) + (a+b)]}{2}
   \]
   \[
   a^{670}(a-b)^{669}(a+b)^{669} = \frac{a^{669}(a-b)^{669}(2a)}{2}
   \]
   \[
   a^{670}(a-b)^{669}(a+b)^{669} = a^{670}(a-b)^{669}(a+b)^{669}
   \]
   This confirms that \( y \) is indeed the average of \( x \) and \( z \).

5. **Check if the product is a perfect 2008th power:**
   Calculate the product \( P \):
   \[
   P = x \cdot y \cdot z = [a^{669}(a-b)^{670}(a+b)^{669}] \cdot [a^{670}(a-b)^{669}(a+b)^{669}] \cdot [a^{669}(a-b)^{669}(a+b)^{670}]
   \]
   Combine the exponents:
   \[
   P = a^{669+670+669} (a-b)^{670+669+669} (a+b)^{669+669+670}
   \]
   \[
   P = a^{2008} (a-b)^{2008} (a+b)^{2008}
   \]
   \[
   P = [a(a-b)(a+b)]^{2008}
   \]
   Since \( a(a-b)(a+b) \) is an integer, \( P \) is indeed a perfect 2008th power.

Conclusion:
\[
\boxed{\text{Yes}}
\]