Compute \[\left\lfloor \dfrac{2007!+2004!}{2006!+2005!}\right\rfloor.\] (Note that $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.)
1. Start with the given expression:
   \[
   \left\lfloor \dfrac{2007! + 2004!}{2006! + 2005!} \right\rfloor
   \]

2. Factor out \(2004!\) from both the numerator and the denominator:
   \[
   \dfrac{2007! + 2004!}{2006! + 2005!} = \dfrac{2004! \cdot (2005 \cdot 2006 \cdot 2007 + 1)}{2004! \cdot (2005 \cdot 2006 + 2005)}
   \]

3. Simplify the expression by canceling \(2004!\):
   \[
   \dfrac{2005 \cdot 2006 \cdot 2007 + 1}{2005 \cdot 2006 + 2005}
   \]

4. Factor out \(2005\) from the denominator:
   \[
   2005 \cdot 2006 + 2005 = 2005 \cdot (2006 + 1) = 2005 \cdot 2007
   \]

5. Substitute this back into the expression:
   \[
   \dfrac{2005 \cdot 2006 \cdot 2007 + 1}{2005 \cdot 2007}
   \]

6. Separate the fraction:
   \[
   \dfrac{2005 \cdot 2006 \cdot 2007}{2005 \cdot 2007} + \dfrac{1}{2005 \cdot 2007}
   \]

7. Simplify the first term:
   \[
   \dfrac{2005 \cdot 2006 \cdot 2007}{2005 \cdot 2007} = 2006
   \]

8. Combine the terms:
   \[
   2006 + \dfrac{1}{2005 \cdot 2007}
   \]

9. Since \(\dfrac{1}{2005 \cdot 2007}\) is a very small positive number, the floor function of the sum will be:
   \[
   \left\lfloor 2006 + \dfrac{1}{2005 \cdot 2007} \right\rfloor = 2006
   \]

Conclusion:
\[
\boxed{2006}
\]