Determine all pairs $(x, y)$ of positive integers for which the equation \[x + y + xy = 2006\] holds.
1. Start with the given equation:
   \[
   x + y + xy = 2006
   \]

2. Add 1 to both sides of the equation:
   \[
   x + y + xy + 1 = 2007
   \]

3. Notice that the left-hand side can be factored:
   \[
   (x + 1)(y + 1) = 2007
   \]

4. Factorize 2007:
   \[
   2007 = 3^2 \cdot 223
   \]

5. List all factor pairs of 2007:
   \[
   2007 = 1 \cdot 2007, \quad 3 \cdot 669, \quad 9 \cdot 223
   \]

6. Each factor pair \((a, b)\) corresponds to \((x+1, y+1)\):
   \[
   (x+1, y+1) = (1, 2007), (2007, 1), (3, 669), (669, 3), (9, 223), (223, 9)
   \]

7. Convert these pairs back to \((x, y)\):
   \[
   (x, y) = (0, 2006), (2006, 0), (2, 668), (668, 2), (8, 222), (222, 8)
   \]

8. Since we are only interested in positive integers, we discard the pairs \((0, 2006)\) and \((2006, 0)\).

9. The remaining pairs are:
   \[
   (x, y) = (2, 668), (668, 2), (8, 222), (222, 8)
   \]

Conclusion:
\[
(x, y) = \boxed{(2, 668), (668, 2), (8, 222), (222, 8)}
\]