Yes. Take $a_{2^k} = \dfrac{1}{2^k}$ and $a_k=0$ if $k$ isn't a power of $2$.

Brunito vos no deberias estar viendo la premiacion de OMA? en fin.. Let the given equation be $$ a_k = \sum_{n=2}^\infty a_{nk} $$We are given $a_1 = 1$. For $k=1$, we have $a_1 = \sum_{n=2}^\infty a_n = 1$. Let $a_k = \frac{A}{k^\alpha}$ for some constants $A, \alpha > 0$. Then we have $$ \frac{A}{k^\alpha} = \sum_{n=2}^\infty \frac{A}{(nk)^\alpha} = \frac{A}{k^\alpha} \sum_{n=2}^\infty \frac{1}{n^\alpha} $$$$ 1 = \sum_{n=2}^\infty \frac{1}{n^\alpha} $$Let $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$ be the Riemann zeta function. Then we require $$ \sum_{n=2}^\infty \frac{1}{n^\alpha} = \zeta(\alpha) - 1 = 1 $$which implies $\zeta(\alpha) = 2$. Since $\zeta(s)$ is a continuous and decreasing function for $s > 1$, and $\lim_{s \to 1^+} \zeta(s) = \infty$ and $\lim_{s \to \infty} \zeta(s) = 1$, there exists a unique $\alpha > 1$ such that $\zeta(\alpha) = 2$. For such $\alpha$, we can choose $a_k = \frac{A}{k^\alpha}$ for any $A > 0$. In particular, we can choose $A = 1$. Therefore, we can choose $a_k = \frac{1}{k^\alpha}$ such that $\zeta(\alpha) = 2$. The value of $\alpha$ such that $\zeta(\alpha) = 2$ is approximately $\alpha \approx 1.728647$. Therefore, such a sequence exists.

We claim such a sequence exists (in fact, the cardinality of the set of all these sequences is $\aleph$). Choose $a_n$ to be a multiplicitive sequence and $\left\{a_n\right\}_{n=1}^\infty\subset\left(0,1\right)$, now the only condition left to satisfy is for the sum of all elements to be 2, since we are summing a multiplicitive series we can use a well known way to turn it to a product: $$2=\sum_{n=1}^\infty a_n=\prod_{\text{p prime}}\left(1+a_p+a_p^2+\dots\right)=\prod_{\text{p prime}}\frac{1}{1-a_p}$$Now there are obviously many many different ways of choosing such $0<a_p<1$ and they define the entire series uniquely, in particular such a sequence exists. For example we can take for the n'th prime $p_n$: $$a_{p_n}=1-\left(\frac{\frac{n+1}{n}}{\frac{n+2}{n+1}}\right)^{-1}=\frac{1}{\left(n+1\right)^2}$$