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.
Claim 1
It suffices to show that the ratio $\frac{S(a)}{N(a)}$ can be made arbitrarily large (rather than actually infinite).
Proof
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 $S(a) = \infty$.
Claim 2
The ratio $\frac{S(a)}{N(a)}$ can be made arbitrarily large.
Proof
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.