TITLE: IMO 1979 problem
QUESTION [11 upvotes]: The question is
$$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$
So my solution went like this:

$$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$
Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$.
Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).

Now my question is:
Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$
So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$.
Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning?
Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution

I realised my mistake after some hours of pondering over it.
(In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this:  $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time} $. It was just a coincidence that the value I calculated was divisible by $1979$.
The correct way, indeed, would be through the link supplied above

REPLY [4 votes]: Here is the solution in the comments, typed out as an answer.
The trick is to separate out the negative terms:
$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{1318}  + \frac{1}{1319}$$
$$= 1 + \frac{1}{2} + \frac{1}{3}   + \frac{1}{1319} - 2 \left( \frac{1}{2} + \frac{1}{4}  + \cdots + \frac{1}{1318} \right)$$
$$=\frac{1}{660} + \frac{1}{661} + \cdots + \frac{1}{1319}.$$
and to notice that $660 + 1319 = 1979$. Combine terms in pairs from the outside:
$$\frac{1}{660} + \frac{1}{1319} = \frac{1979}{660 \cdot 1319}$$
$$\frac{1}{661} + \frac{1}{1318} = \frac{1979}{661 \cdot 1318}$$
$$\cdots  =\cdots$$
and so on.
There are an even number of terms, so this gives us a sum of terms in the form $\frac{1979}{m}$, with m not divisible by $1979$ (since $1979$ is prime and so does not divide any product of smaller numbers).
Hence the sum of the terms gives a rational number with denominator not divisible by $1979$. Q.E.D