Is there any sequence $a_n$ of nonnegative numbers for which $\displaystyle\sum_{n \geq 1}a_n^2 <\infty$ and
$$\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty\quad?$$
See also https://math.stackexchange.com/questions/42624/double-sum-miklos-schweitzer-2010
Yes, such sequences exist.
In effect the problem concerns the linear operator, call it $T$, that maps any sequence $(a_n)$ to the sequence whose $n$-th term is $\sum_{k\geq1} a_{kn}/k$. The problem asks whether there exists $a$ such that the $l^2$ norm $\|a\|_2$ is finite but $\|Ta\|_2 = \infty$. We show that such $a$ exist for each $l^q$ norm ($q \geq 1$). [I can't call it $l^p$ because I'll soon want to use $p$ for a generic prime number.]
The key is that $T$ can be regarded as an infinite tensor product (Kronecker product) of operators $T_p$, where for each prime $p$ the operator $T_p$ is convolution with $(1,p^{-1},p^{-2},p^{-3},\ldots)$ on $(a_1,a_p,a_{p^2},a_{p^3},\ldots)$. The operator $T_p$ has norm $\sum_{i=0}^\infty p^{-i} = p/(p-1)$, so $T = \otimes_p T_p$ should have norm $\prod_p p/(p-1) = \infty$.
To get a construction from this, we first make for each $M>1$ a nonnegative sequence $\alpha = (\alpha_n)$ such that $\|\alpha\|_q < \infty$ and $T\alpha = M\alpha$. Since $\prod_p p/(p-1)$ diverges, we can find $x$ large enough that $\prod_{p\leq x} p/(p-1) > M$. Choose for each prime $p\leq x$ a positive number $z_p$ smaller than 1 but close enough to 1 that $\prod_{p\leq x} p/(p-z_p) = M$. Now let $\alpha$ be the totally multiplicative function taking each $p\leq x$ to $z_p$ and each $p>x$ to $0$. (That is, if $n=\prod_{p\leq x} p^{e_p}$ then $\alpha_n = \prod_{p \leq x} z_p^{e_p}$, while if $n$ has a prime factor bigger than $x$ than $\alpha_n = 0$.) Then the $n$-th term of $T\alpha$ is $$ \sum_{k\geq 1} \frac{\alpha_{kn}}{k} = \sum_{k\geq 1} \frac{\alpha_k \alpha_n}{k} = \alpha_n \sum_{k\geq 1} \frac{\alpha_k}{k} $$ because $\alpha$ is multiplicative, and $\sum_{k\geq 1} \alpha_k/k$ is the product over $p\leq x$ of geometric series $$ \sum_{e=0}^\infty \frac{\alpha_{p^e}}{p^e} = \sum_{e=0}^\infty (z_p/p)^e = \frac{p}{p-z_p}, $$ so $$ T\alpha = \prod_{p\leq x} \frac{p}{p-z_p} \cdot \alpha = M \alpha; $$ and $\|\alpha\|_q < \infty$ because $\|\alpha\|_q^q$ is the product of $\pi(x)$ convergent geometric series.
The conclusion is a standard argument (as already noted in the parallel stackexchange thread). For example, let $a^{(M)} = \alpha / \|\alpha\|$ where $\alpha$ is a sequence with $T\alpha = M\alpha$ as in the last paragraph. Then $\|a^{(M)}\| = 1$ and $Ta^{(M)} = M a^{(M)}$. Set $$ a = \sum_{m=1}^\infty \frac{a^{(4^m)}}{2^m} = \frac{a^{(4)}}{2} + \frac{a^{(16)}}{4} + \frac{a^{(64)}}{8} + \cdots . $$ The $m$-th term has norm $1/2^m$, so the sum converges in $l^q$ to a vector of norm at most $\sum_{m=1}^\infty 1/2^m = 1$. But since each term $a^{(4^m)} / 2^m$ is a nonnegative sequence, and $T$ has nonnegative coefficients, we have for each $m \geq 1$ $$ Ta > T\frac{a^{(4^m)}}{2^m} = 2^m a^{(4^m)}; $$ and $2^m a^{(4^m)}$ is a vector of norm $2^m$. Hence $\|Ta\|_q = \infty$, QED.
In the recent issue of Matematikai Lapok, the report on the Schweitzer contest attributes the problem to Zoltan Buczolich and Julien Bremont.
Solution (based on Buczolich Zoltán's solution): There exists such a sequence. For every $M \in \mathbb{N}$, choose a set $\mathcal{P}_M = \left\{ p_{j, M} : j = 1, \ldots, l_M \right\}$ of distinct primes. Ensure that these sets are disjoint for different values of $M$. Additionally, exploiting the fact that
$\sum_{\text{prime } p} \frac{1}{p} = \infty,$
we choose the sets $\mathcal{P}_M$ such that
$$\sum_{j=1}^{l_M} \frac{1}{p_{j, M}} > \sqrt{2} M.$$
Next, choose $m_M$ such that
$$\frac{(m_M - 1)^{l_M}}{m_M^{l_M}} > \frac{1}{2}.$$ Let
$$N_M = \left\{p_{1, M}^{a_1} \cdots p_{l_M, M}^{a_{l_M}}: 1 \leq a_j \leq m_M, j = 1, \ldots, l_M\right\},$$
and
$$N_M' = \left\{p_{1, M}^{a_1} \cdots p_{l_M, M}^{a_{l_M}}: 1 \leq a_j \leq m_M - 1, j = 1, \ldots, l_M\right\}.$$
The sets $N_M$ are disjoint for different values of $M$, with $\#N_M = m_M^{l_M}$ and $\#N_M' = (m_M - 1)^{l_M}$.
If $n \in N_M$, define $a_n = \frac{1}{M \sqrt{m_M^{l_M}}}$. Otherwise, if $n \notin \cup_M N_M$, let $a_n = 0$. It is easy to see that $(a_n) \in \ell_2$. Furthermore,
$$\sum_{n \geq 1}\left(\sum_{k \geq 1} \frac{a_{kn}}{k}\right)^2 \geq \sum_{M=1}^{\infty} \sum_{n \in N_M'}\left(\sum_{k \geq 1} \frac{a_{kn}}{k}\right)^2$$
$$\geq \sum_{M=1}^{\infty} \sum_{n \in N_M'}\left(\sum_{k \in \mathcal{P}_M} \frac{a_{kn}}{k}\right)^2 = \sum_{M=1}^{\infty} \sum_{n \in N_M'} \frac{1}{M^2 m_M^{l_M}} \left(\sum_{j=1}^{l_M} \frac{1}{p_{j, M}}\right)^2$$
$$\geq \sum_{M=1}^{\infty} \sum_{n \in N_M'} \frac{2}{m_M^{l_M}} = \sum_{M=1}^{\infty} \frac{2(m_M - 1)^{l_M}}{m_M^{l_M}} \geq \sum_{M=1}^{\infty} 1 = \infty.$$
Thus, no correct solution has been found for this problem.
