Determine which integers $n>1$ have the property that there exists an infinite sequence $a_{1}, a_{2}, a_{3}, \ldots$ of nonzero integers such that the equality
$$
a_{k}+2 a_{2 k}+\cdots+n a_{n k}=0
$$
holds for every positive integer $k$.
Answer: all $n>2$.
For $n=2$, we have $a_{k}+2 a_{2 k}=0$, which is clearly not possible, since it implies $a_{2^{k}}=\frac{a_{1}}{2^{k-1}}$ for all $k \geq 1$

For $n \geq 3$ we will construct a completely multiplicative sequence (meaning $a_{i j}=a_{i} a_{j}$ for all $i$ and $j$ ). Thus $\left(a_{i}\right)$ is determined by its value on primes, and satisfies the condition as long as $a_{1}+2 a_{2}+\cdots+n a_{n}=0$. The idea is to take two large primes and use Bezout's theorem, but the details require significant care.

We start by solving the case where $n \geq 9$. In that case, by Bertrand postulate there exists primes $p$ and $q$ such that
$$
\lceil n / 2\rceil3$. Also, $pn$, and $4 p \geq 4\left(\frac{1}{2}(q+1)\right)>$ $n$. We now stipulate that $a_{r}=1$ for any prime $r \neq p, q$ (in particular including $r=2$ and $r=3$ ). There are now three cases, identical in substance.
- If $p, 2 p, 3 p \in[1, n]$ then we would like to choose nonzero $a_{p}$ and $a_{q}$ such that
$$
6 p \cdot a_{p}+q \cdot a_{q}=6 p+q-\frac{1}{2} n(n+1)
$$
which is possible by Bézout lemma, since $\operatorname{gcd}(6 p, q)=1$.
- Else if $p, 2 p \in[1, n]$ then we would like to choose nonzero $a_{p}$ and $a_{q}$ such that
$$
3 p \cdot a_{p}+q \cdot a_{q}=3 p+q-\frac{1}{2} n(n+1)
$$
which is possible by Bézout lemma, since $\operatorname{gcd}(3 p, q)=1$.
- Else if $p \in[1, n]$ then we would like to choose nonzero $a_{p}$ and $a_{q}$ such that
$$
p \cdot a_{p}+q \cdot a_{q}=p+q-\frac{1}{2} n(n+1)
$$
which is possible by Bézout lemma, since $\operatorname{gcd}(p, q)=1$. (This case is actually possible in a few edge cases, for example when $n=9, q=7, p=5$.)

It remains to resolve the cases where $3 \leq n \leq 8$. We enumerate these cases manually:
- For $n=3$, let $a_{n}=(-1)^{\nu_{3}(n)}$.
- For $n=4$, let $a_{n}=(-1)^{\nu_{2}(n)+\nu_{3}(n)}$.
- For $n=5$, let $a_{n}=(-2)^{\nu_{5}(n)}$.
- For $n=6$, let $a_{n}=5^{\nu_{2}(n)} \cdot 3^{\nu_{3}(n)} \cdot(-42)^{\nu_{5}(n)}$.
- For $n=7$, let $a_{n}=(-3)^{\nu_{7}(n)}$.
- For $n=8$, we can choose $(p, q)=(5,7)$ in the prior construction.

This completes the constructions for all $n>2$.