solution: 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{align*}
  5^0 &= 1 \\
  5^1 &= 5 \\
  5^2 &= 25 \\
  5^3 &= 125 \\
  5^4 &= 625
  \end{align*} 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 $1+201 = \boxed{202}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  If $\frac{a}{b}$ is the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal, with $a$ and $b$ being relatively prime positive integers, what is $a+b$?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 202
extra_info.question: If $\frac{a}{b}$ is the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal, with $a$ and $b$ being relatively prime positive integers, what is $a+b$?
extra_info.split: test