Let $ \alpha_1$, $ \alpha_2$, $ \ldots$, $ \alpha_{2008}$ be real numbers. Find the maximum value of \[ \sin\alpha_1\cos\alpha_2 \plus{} \sin\alpha_2\cos\alpha_3 \plus{} \cdots \plus{} \sin\alpha_{2007}\cos\alpha_{2008} \plus{} \sin\alpha_{2008}\cos\alpha_1\]
1. Consider the given expression:
   \[
   S = \sin\alpha_1\cos\alpha_2 + \sin\alpha_2\cos\alpha_3 + \cdots + \sin\alpha_{2007}\cos\alpha_{2008} + \sin\alpha_{2008}\cos\alpha_1
   \]

2. We use the trigonometric identity for the product of sine and cosine:
   \[
   \sin x \cos y = \frac{1}{2} [\sin(x + y) + \sin(x - y)]
   \]

3. Applying this identity to each term in the sum, we get:
   \[
   \sin\alpha_i \cos\alpha_{i+1} = \frac{1}{2} [\sin(\alpha_i + \alpha_{i+1}) + \sin(\alpha_i - \alpha_{i+1})]
   \]

4. Substituting this into the original sum, we have:
   \[
   S = \frac{1}{2} \sum_{i=1}^{2008} [\sin(\alpha_i + \alpha_{i+1}) + \sin(\alpha_i - \alpha_{i+1})]
   \]

5. Notice that the terms \(\sin(\alpha_i - \alpha_{i+1})\) can be positive or negative, but their sum is bounded by the number of terms, which is 2008. The maximum value of \(\sin(\alpha_i + \alpha_{i+1})\) is 1, and similarly for \(\sin(\alpha_i - \alpha_{i+1})\).

6. To maximize \(S\), we need to maximize the sum of \(\sin(\alpha_i + \alpha_{i+1})\). The maximum value of \(\sin(\alpha_i + \alpha_{i+1})\) is 1, and this occurs when \(\alpha_i + \alpha_{i+1} = \frac{\pi}{2}\) or \(\alpha_i - \alpha_{i+1} = \frac{\pi}{2}\).

7. If we set \(\alpha_1 = \alpha_2 = \cdots = \alpha_{2008} = \frac{\pi}{4}\), then:
   \[
   \sin\alpha_i = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
   \]
   and
   \[
   \cos\alpha_i = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
   \]

8. Substituting these values into the original sum, we get:
   \[
   S = \sum_{i=1}^{2008} \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}\right) = \sum_{i=1}^{2008} \left(\frac{1}{2}\right) = 1004
   \]

9. Therefore, the maximum value of the given expression is:
   \[
   \boxed{1004}
   \]