But it is 1000 times!

We want to find the number of positive integers $n$ such that $\dfrac{2003n+2002}{20n+2}=m\in \mathbb{Z}$ Simple polynomial division gives: $\begin{aligned} m=\dfrac{2003n+2002}{20n+2}&=\dfrac{2003}{20}+\dfrac{1801.7}{20n+2}\\ \implies 20m &=2003+\dfrac{18017}{10n+1} \end{aligned}$ On factorizing, $18017=43\times 419$ .Therefore, it's only factors are $1,43,419,18017$ .None of these is of the form $10n+1$ .Hence there is no positive integer $n$ which will make $\dfrac{2003n+2002}{20n+2}$ and consequentially $m$ an integer.

Let $Q = \dfrac {2003n + 20002}{20n + 2} = \dfrac {100(20n+2)+1802}{20n+2} = 100 + \dfrac {1802}{20n+2}$ . For $Q$ to be a positive integer $\dfrac {1802}{20n+2} = \dfrac {2\times 17 \times 53}{\color{#3D99F6}20n + 2} = \dfrac {2\times 17 \times 53}{\color{#3D99F6} 2 \times (10n +1)} = \dfrac {17 \times 53}{\color{#3D99F6} 10n +1}$ must be a positive integer.

It is simple to notice that there is no value for n which would satisfy the conditions, hence the answer is $\boxed{0}$ .

Disclaimer: This question seems similar to my question

A simple question which can be solved by polynomial division and knowing the fact that if number is divided by its factor the remainder is zero.The given dividend is 2003n +2002 and divisor is 20n+2. On dividing the dividend by divisor the remainder obtained is 3n+1802. Since the question asks us to find such a "n" such that the divisor is factor of the dividend, which implies the remainder = 0, 3n + 1802 =0, => n= -600.66667 . But the question asks for the number of positive integers and our answer is negative as well as not an integer. Hence, clearly the answer is zero.....