One competition in Macedonia 1987 ( ex YUG )

[rsoldo]

Prove that for any choice of the sign $+$ or $-$ in the sum

$$1\pm\frac{1}{2}\pm\frac{1}{3}\pm\ldots\pm\frac{1}{11}\pm\frac{1}{12}$$

we can never get $0$.

[nnosipov]

I suppose, we can use that there is $\pm 1/2^3$ in the sum. So, we can't obtain an integer as the value of sum (not only zero).

[svv]

Допустим, равенство возможно. Тогда
$\pm 1\pm\frac{1}{2}\pm\ldots\pm\frac{1}{10}\pm\frac{1}{12}=\frac{1}{11}$
В левой части приведём дроби к общему знаменателю и сложим их. Знаменатель должен быть в $11$ раз больше числителя, но он не содержит множителя $11$.

[mihaild]

We can also use $\frac{1}{11}$, $\frac{1}{9}$ or $\frac{1}{7}$ instead - take any fraction s.t. it's denominator doesn't divide lcm of others' denominators, multiply all fractions by this lcm, and all but one fractions become integers, so even after this multiplication sum isn't integer.

[nnosipov]

This is the same idea. We have some prime number $p$ which divides (in its maximal exponent) only one of the denominators. Then the sum cannot be integer. One can be used any $p \in \{2,3,5,7,11\}$.

Upd. $p=5$ doesn't work because we have two denominators dividing by $5$.

Upd-2. In fact, $p=5$ works: it suffices to add $\pm 1/5$ and $\pm 1/10$.