jmc

Let $n=2004,$ so the expression becomes $$\lfloor \frac{(n+1{)}^3}{(n-1)n}-\frac{(n-1{)}^3}{n(n+1)}\rfloor .$$Combining the fractions under the common denominator $(n-1)n(n+1),$ we get $$\begin{array}{rl}\lfloor \frac{(n+1{)}^3}{(n-1)n}-\frac{(n-1{)}^3}{n(n+1)}\rfloor & =\lfloor \frac{(n+1{)}^4-(n-1{)}^4}{(n-1)n(n+1)}\rfloor \\ & =\lfloor \frac{(n^4+4n^3+6n^2+4n+1)-(n^4-4n^3+6n^2-4n+1)}{(n-1)n(n+1)}\rfloor \\ & =\lfloor \frac{8n^3+8n}{(n-1)n(n+1)}\rfloor \\ & =\lfloor \frac{8n(n^2+1)}{(n-1)n(n+1)}\rfloor \\ & =\lfloor \frac{8(n^2+1)}{n^2-1}\rfloor .\end{array}$$Because $\frac{n^2+1}{n^2-1}$ is a little greater than $1,$ we expect $\frac{8(n^2+1)}{n^2-1}$ to be a little greater than $8,$ which makes the floor equal to $\boxed{8}.$

Indeed, we have $$\frac{n^2+1}{n^2-1}=1+\frac{2}{n^2-1}=1+\frac{2}{2004^2-1}<1+\frac{2}{1000}=1.002,$$so $\frac{8(n^2+1)}{n^2-1}<8.016,$ so $8<\frac{8(n^2+1)}{n^2-1}<8.016<9,$ as claimed.