jmc

There are 2009 positive integers less than 2010, of which 1005 are odd. If $\frac{1}{n}$ is equal to a terminating decimal, then $n$ can only be divisible by 2 and 5. However, since we have the added restriction that $n$ is odd, $n$ must be a power of 5. There are five powers of 5 less than 2010. $$\begin{array}{rl}{5}^0& =1\\ 5^1& =5\\ 5^2& =25\\ 5^3& =125\\ 5^4& =625\end{array}$$ Note that $5^5=3125$. Since there are five odd integers that satisfy our desired condition, the desired probability is $\frac{5}{1005}=\frac{1}{201}$. This is in simplest terms, so our answer is $1+201=\boxed{202}$.