I came up with this problem on a book in number theory :
Let $p$ and $q$ be natural numbers such that $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{1335} = \frac{p}{q} \,.$$ Show that $2003 \mid p$.
Any ideas of how to start solving it? Suggestions and helps would be greatly appreciated!
Suppose that $k>1$ is a positive integer such that $3k-1$ is a prime number. Then the numerator of the alternating series $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{2k-1}$$ is divisible by $3k-1$. For example when $k=2$, we have $2k-1=3$ and $$1-\frac12+\frac13=\frac{5}{6}$$ with numerator divisible by $5=3k-1$. For $k=4$, we have $2k-1=7$ and $3k-1=11$, and $11$ divides the numerator of $$1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17=\frac{319}{420}.$$ The claim may not hold if $3k-1$ is not prime. For example if $k=3$, we have $3k-1=8$, and $8$ doesn't divide the numerator of $$1-\frac12+\frac13-\frac14+\frac15=\frac{47}{60}.$$
The OP's problem comes from $k=668$, where $2k-1=1335$ and $3k-1=2003$. Here is a proof of this claim.
Let $S$ denote the series to be evaluated. Then, $$S=\left(1+\frac12+\frac13+\frac14+\ldots+\frac{1}{2k-1}\right)-2\left(\frac12+\frac14+\ldots+\frac{1}{2k-2}\right).$$ Hence $$S=\left(1+\frac12+\frac13+\frac14+\ldots+\frac{1}{2k-1}\right)-\left(1+\frac12+\ldots+\frac{1}{k-1}\right).$$ This makes $$S=\frac1k+\frac1{k+1}+\ldots+\frac1{2k-2}+\frac{1}{2k-1}.$$ Note that $3k-1$ must be odd, so $k$ is even. That is, $$S=\left(\frac1k+\frac1{2k-1}\right)+\left(\frac{1}{k+1}+\frac{1}{2k-2}\right)+\ldots+\left(\frac{1}{\frac{3}{2}k-1}+\frac{1}{\frac{3}{2}k}\right).$$ That is $$S=\frac{3k-1}{k(2k-1)}+\frac{3k-1}{(k+1)(2k-2)}+\ldots+\frac{3k-1}{\left(\frac{3}{2}k-1\right)\frac32k}.$$ Thus $$S=(3k-1)\left(\frac{1}{k(2k-1)}+\frac{1}{(k+1)(2k-2)}+\ldots+\frac{1}{\left(\frac{3}{2}k-1\right)\frac32k}\right).$$ Because $3k-1$ is a prime greater than $k,k+1,\ldots,2k-2,2k-1$, it is not cancelled out from the numerator of $S$.
I think it is an interesting problem to see whether there exists an integer $k>1$ such that $3k-1$ is not prime but $3k-1$ divides the numerator of $S$. I conjecture that such $k$ don't exist.