Given that $ x \plus{} \sin y \equal{} 2008$ and $ x \plus{} 2008 \cos y \equal{} 2007$, where $ 0 \leq y \leq \pi/2$, find the value of $ x \plus{} y$.
1. We start with the given equations:
   \[
   x + \sin y = 2008
   \]
   \[
   x + 2008 \cos y = 2007
   \]

2. Subtract the second equation from the first equation:
   \[
   (x + \sin y) - (x + 2008 \cos y) = 2008 - 2007
   \]
   Simplifying this, we get:
   \[
   \sin y - 2008 \cos y = 1
   \]

3. Since \(0 \leq y \leq \frac{\pi}{2}\), we know that \(\cos y\) is strictly positive and \(\sin y\) ranges from 0 to 1. We analyze the equation \(\sin y - 2008 \cos y = 1\):
   \[
   \sin y = 1 + 2008 \cos y
   \]

4. Given that \(\sin y \leq 1\), we substitute this into the equation:
   \[
   1 \geq 1 + 2008 \cos y
   \]
   Simplifying, we get:
   \[
   0 \geq 2008 \cos y
   \]
   Since \(\cos y\) is positive, the only way this inequality holds is if \(\cos y = 0\).

5. If \(\cos y = 0\), then:
   \[
   \sin y = 1
   \]
   This implies:
   \[
   y = \frac{\pi}{2}
   \]

6. Substitute \(y = \frac{\pi}{2}\) back into the first equation to find \(x\):
   \[
   x + \sin\left(\frac{\pi}{2}\right) = 2008
   \]
   \[
   x + 1 = 2008
   \]
   \[
   x = 2007
   \]

7. Finally, we find \(x + y\):
   \[
   x + y = 2007 + \frac{\pi}{2}
   \]

The final answer is \(\boxed{2007 + \frac{\pi}{2}}\).