Question 11
In a sum there are $1669$ fractions and the denominators rearrange the integers from $1$ to $1669$. If the denominator is a multiple of $5$, then the numerator is $4$ and the fraction is signed $-$; If the denominator is not a multiple of $5$, then the numerator is $1$ and the fraction is signed $+$:
$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{4}{5} + \dots + \frac{1}{1661} + \frac{1}{1662} + \frac{1}{1663} + \frac{1}{1664} - \frac{4}{1665} + \frac{1}{1666} + \frac{1}{1667} + \frac{1}{1668} + \frac{1}{1669}
$$
If the sum is written as an irreducible (simplified) fraction, show that the numerator of this fraction is divisible by $2003$.
$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{4}{5} + \dots + \frac{1}{1661} + \frac{1}{1662} + \frac{1}{1663} + \frac{1}{1664} - \frac{4}{1665} + \frac{1}{1666} + \frac{1}{1667} + \frac{1}{1668} + \frac{1}{1669}
$$
If the sum is written as an irreducible (simplified) fraction, show that the numerator of this fraction is divisible by $2003$.