Show that $2003$ divides the numerator of $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{1335}$. 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 $k&gt;1$ 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 $k&gt;1$ such that $3k-1$ is not prime but $3k-1$ divides the numerator of $S$.  I conjecture that such $k$ don't exist.
COMMENT:

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ . . .+ \frac{1}{1335}=\frac{p}{q}$$

⇒$$\frac{ 1335 t +1334!}{1335!}=\frac{p}{q}$$

If $2003|p$, then we must have:

$1335 t+1334!=2003 k$

Now consider equation $1335 t +1=2003 k_1$

One solution of this equation is:

$(t, k_1)=(3, 2)$

Now we have to show that $p=1335 t+1334! ≡1335 t +1$, or:

$1334! ≡ 1 \mod (2003)$