IMO Problem-year[ $p$ ]

In this solution, I have approached to find a prime which divides $a$ but not $b$ :

Let $\large S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{1318}+\frac{1}{1319}-2(\frac{1}{2}+\frac{1}{4}+\cdot+\frac{1}{1318})$

$\large\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{1318}+\frac{1}{1319}-\frac{2}{2}(1+\frac{1}{2}+\cdot+\frac{1}{659})$

$\large\Rightarrow S=\frac{1}{660}+\frac{1}{661}+\frac{1}{662}+\cdots+\frac{1}{1318}+\frac{1}{1319}$

Now observe that $660+1319$ is a prime number i.e. $1979$

So $\large\frac{1}{660}+\frac{1}{1319}=\frac{1979}{660\cdot1319}; \frac{1}{661}+\frac{1}{1318}=\frac{1979}{661\cdot1318}$ and so on...

So taking out $1979$ as common we get $\large S=1979(\frac{1}{660\cdot1319}+\frac{1}{661\cdot1318}\cdots\frac{1}{989\cdot990})$

Since the $1979$ is prime it doesn't divide the denominator. So the answer is $\boxed{1979}$

I searched through the first 1000 primes using $Mathematica$ and got it right

Select[Prime@Range@1000,Mod[Numerator@Sum[(-1)^(n+1)*1/n,{n,1319}],#]==0&]

this code returns $1979$