There is a question in the Miklos Schweitzer contest last year that keeps bugging me. Here it is:
Is there any sequence $(a_n)$ of nonnegative numbers for which $\displaystyle\sum_{n \geq 1}a_n^2 <\infty $ and $$\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty\quad?$$
I figured out something of potential interest, but I am uncertain of its usefulness.
Also, I apparently can't comment yet because I don't have enough privilege points or something, so I'm "answering". (The scratchwork wouldn't fit in a comment anyway.)
Define $ \sum a_n^2 = L $ and note that $ n^2 \le \sigma_2(n) < \zeta(2) n^2 $ (this can be seen by factoring $ \sigma_2(n)n^{-2} $ and then noting that it is always a finite part of the Euler product for the Riemann zeta function but can allow arbitrarily many of its terms). Use $ (n,m) $ for greatest common divisor. Then
$$ S = \sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2 $$ $$ = \sum_{n,m=1}^\infty \left( \sum_{d|(n,m)} \frac{1}{(n/d)(m/d)} \right) a_n a_m $$ $$ = \sum_{n,m=1}^\infty \sigma_2((n,m)) \frac{a_n}{n} \frac{a_m}{m} $$ $$ = \sum_{n=1}^\infty \sigma_2 (n) (a_n / n)^2 + 2 \sum_{n=2}^\infty \left( \sum_{m<n} \frac{\sigma_2((n,m))}{m} a_m \right) \frac{a_n}{n} $$ $$ <\zeta(2) \left( L + 2 \sum_{n=2}^\infty \left( \sum_{m<n} \frac{(n,m)^2}{m} a_m \right) \frac{a_n}{n} \right) $$
Hence if we can prove $$ M = \sup \left\{ \frac{1}{n a_n} \sum_{m<n} \frac{(n,m)^2}{m} a_m \right\} < \infty $$ we may then establish the existence of finite upper bound
$$ S < \zeta(2) (L + 2M (L-a_1^2)) .$$
I'd look into this more (namely on how to find a lower bound practical enough for the approach from the other side) but it's the middle of the night here. Maybe later.
(This is not an answer.) The situation is even worse than Patrick Da Silva suggests. Let $a_n = \frac{b_n}{\sqrt{n}}$; then $\sum \frac{b_n^2}{n}$ converges, so $\liminf b_n = 0$. Moreover $\frac{a_{nk}}{k} = \frac{b_{nk}}{k^{3/2} \sqrt{n}}$, hence the above sum gives
$$\sum_{n \ge 1} \frac{1}{n} \left( \sum_{k \ge 1} \frac{b_{nk}}{k^{3/2}} \right)^2.$$
In particular, if $b_n$ is eventually monotonically decreasing (or even some weaker form of this assumption), then the sum in parentheses is eventually at most $b_n^2 \zeta \left( \frac{3}{2} \right)$ and so $a_n$ cannot be a counterexample to the given statement.
In other words, a counterexample needs to be fairly non-monotonic (if the statement is false). It seems like a good idea to make $a_n$ large if $n$ has many factors, but I haven't been able to do anything concrete with this.
I thought I had something, but as ShreevatsaR correctly remarked, I made a stupid mistake, and the below proof is false. However, Claim 1 is (I hope) correct, and perhaps the idea beyond claim 2 can be used by someone to produce a proof, so I will not delete the answer (at least for the time being).
Firstly, let $N(a) := \sum_n a_n^2, \lVert a \rVert := \sqrt{N(a)}$ and $S(a) := \sum_n (\sum_k \frac{a_{nk}}{k} )^2$ where $a$ is a sequence of nonnegative numers. We want to prove that $S(a)$ can be infinite while $N(a)$ is finite.
It suffices to show that the ratio $\frac{S(a)}{N(a)}$ can be made arbitrarily large (rather than actually infinite).
Suppose that the ratio $\frac{S(a)}{N(a)}$ can be made as large as we like. Then we can find, for any $M$ a sequence $a^M$ be a sequence with $N(a^M) < \frac{1}{2^{2M}}$ (so that $ \lVert a^M \rVert < \frac{1}{2^M}$) and $S(a^M) > M$. Now, define $a := \sum a^M$. It is a well defined and square-sumable sequence, since $\lVert \cdot \rVert$ is actually a norm. Now, $S(a) \geq S(a^M) > M$ for any $M$, since $a_n \geq a^M_n$ for any $n$ and all coefficients are nonnegative. Thus, we conclude that $S(a) = \infty$.
The ratio $\frac{S(a)}{N(a)}$ can be made arbitrarily large.
Fix an integer $M$ and consider the sequence $a_n = n[n \leq M]$ (which means $a_n = n$ for $n \leq M$ and $a_n = 0$ for $n > M$). We can compute: $$N(a) = \sum_{n=1}^M n^2 $$ and $S(a) = \sum_{n=1}^M (\sum_{k=1}^M \frac{nk}{k})^2 = \sum_{n=1}^M n^2 \cdot M^2 = M^2 N(a)$ Thus, the desired ratio is: $$\frac{S(a)}{N(a)} = M^2$$ which is obviously sufficiently large is $M$ is sufficiently large.
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.
