jmc

We can re-write the product as $$\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots \left(1+\frac{1}{n}\right)=\frac{3}{2}\cdot \frac{4}{3}\cdots \frac{n+1}{n}.$$The numerator of each fraction nicely cancels with the denominator of the next fraction, so the entire product telescopes, leaving just the fraction $\frac{n+1}{2}$. For $\frac{n+1}{2}$ to be an integer, it follows that $n+1$ must be even and $n$ must be odd. The odd numbers from $2\le n\le 2010$ are given by $3,5,\dots ,2009$; there are $\frac{2009-3}{2}+1=\boxed{1004}$ such numbers.