Prove that $a_{n}=0$ for all $n$, if $\sum a_{kn}=0$ for all $k\geq 1$

I assume that your sequence starts at $n=1$

Hint: Notice that we can isolate each term as $$a_k= \sum_{m=1}^\infty \left( \mu(m) \sum_{n=1}^\infty a_{knm} \right)$$ where $\mu$ is the Möbius mu function. (and the index $a_{mnk}$ is the product of variables $k,m,n$) Careful, the above sum on the right hand side might not be absolutely convergent, so an argument with partial sums is needed to show why it is equal to $a_k$.

Hope that helps,

Here's a probabilistic approach.

Define two finite measures $\mu$ and $\nu$ on $\mathbb{Z}_+$ by $\mu(\lbrace n\rbrace)=a_n^+$ and $\nu(\lbrace n\rbrace)={a_n}^-$; the positive and negative parts of $a_n$. The zero sums in the problem mean that $\mu(k\mathbb{Z}_+)=\nu(k\mathbb{Z}_+)$ for every $k\geq 1$.

The collection $\lbrace k\mathbb{Z}_+ : k\geq 1\rbrace$ is a $\pi$-system (closed under finite intersections) and generates the discrete $\sigma$-field on $\mathbb{Z}_+$. The standard application${}^*$ of Dynkin's $\pi,\lambda$ systems shows that $\mu=\nu$. In particular,
$$\mu(\lbrace n\rbrace)=a_n^+ = {a_n}^-=\nu(\lbrace n\rbrace)$$ for all $n$, that is, $a_n=0$ for all $n\geq 1$.

${}^*$ For example, Lemma 1.17 on page 9 of Kallenberg's Foundations of Modern Probability.

Here's a suggestion.. try "sieving" to first show $a_1 = 0$. You first subtract off $\sum a_{2n} = 0$, and get that the sum of $a_n$ over $n$ odd is zero. Then subtract off $\sum a_{3n}$ and you get the resulting series is zero. But now you've subtracted off each $a_{6n}$ twice, so you add back $\sum a_{6n}$. Thus you've now removed all $a_{2n}$ and $a_{3n}$ from your series. You can then proceed with removing all $a_{5n}$, then adding back all $a_{10n}$ and all $a_{15n}$, then subtracting off all $a_{30n}$ to remove all duplicates. The end result is that you've removed all $a_{2n}$, $a_{3n}$, and $a_{5n}$. If you keep doing this, you'll end out removing everything except $a_1$, which therefore is zero.

Then repeat the process to show $a_2 = 0$, first removing off $a_{3n}$, then all $a_{4n}$ and so on. Keep going in this way, until all $a_n$ are shown to be zero.

The only issue I see here is making sure you can always get rid of the duplicates, so this is why I'm just calling this as a suggestion... but it sounds plausible to me.

Let $A(p_1, \dots, p_m)$ be the set of all integers that are divisible by the first $m$ primes $p_1, \dots, p_m$.

I claim that $$S(m) = \sum_{n \in N \setminus A(p_1, \dots, p_m)} x_n = 0.$$

If this is true, then $$S(m) = x_1 + \sum_{n \in A} x_n = 0$$ where $A \subseteq \{p_{m+1}, p_{m+1}+1, \dots\}$. That is, $$|x_1| = \left| \sum_{x \in A} x_n \right| \leq \sum_{n \geq p_m}|x_n|.$$ Because $\sum |x_n|$ converges, if we let $m \to \infty$, we get $x_1=0$.

Next we realize that the sequence $x_{kn}$ satisfies the same conditions as $x_n$. Hence the same reasoning can be applied as above to conclude that $x_{1 \cdot k} = 0$. From this we conclude the entire sequence is zero identically.

It remains to show $$S(m) = \sum_{n \in N \setminus A(p_1, \dots, p_m)} x_n = 0.$$ Because $\sum_n x_n = 0$, this is equivalent to showing $$S(m) = \sum_{n \in A(p_1, \dots, p_m)} x_n = 0.$$

Using inclusion exclusion formula,

$$S(m) = \sum_{i=1}^m \left( (-1)^{i-1} \sum_{I \subseteq \{p_1, \dots, p_m\}, |I|=k} \sum_{n \in A(I)} x_n \right)$$

Where the innermost sum is over all numbers that are multiples of a subset of the primes in $\{p_1, \dots, p_m\}$. But this sum is of the form $\sum_{n} x_{nk}$ for $k$ the product of the primes in the subset, by assumption our sum is zero. Hence the entire expression for $S(m)$ is zero. This proves our claim, which then is used to prove our initial claim.