$2006\geq n\geq \sqrt[3]{500^2}$ and check all the cases...

wow, there will be many cases

The sum equals the cardinal of the set $\{(x,y)\in\mathbb{N}^2\,|\,3n<xy\leq 4n,\, x\geq xy-3n\}$, but we can count how many elements have $y$ equal to $1, 2, 3,\dots$, rewriting the sum in the form of the telescoping series \[ \left(\left[\frac{4n}1\right]-\left[\frac{3n}1\right]\right)+ \left(\left[\frac{4n}2\right]-\left[\frac{3n}2\right]\right)+ \left(\left[\frac{4n}3\right]-\left[\frac{3n}3\right]\right)+ \left(\left[\frac{4n}4\right]-\left[\frac{3n}4\right]\right)+ \left(\left[\frac{3n}4\right]-\left[\frac{3n}5\right]\right)+ \left(\left[\frac{3n}5\right]-\left[\frac{3n}6\right]\right)+ \left(\left[\frac{3n}6\right]-\left[\frac{3n}7\right]\right)+ \cdots \] All that remains is to solve the equation $3n-\left[\frac n2\right]+\left[\frac n3\right]=2006$, but this inmediately implies $\frac{2006-1}{\frac{17}6}<n<\frac{2006+1}{\frac{17}6}$, and so $n=708$.