Does there exist a sequence $\{a_n\}_{n \ge 0}$ of nonnegative reals such that $ \sum_{j \ge 0} a_{nj} = \frac{1}{n}$ holds for all naturals $n$? Does there exist a sequence $\{a_n\}_{n \ge 0}$ of nonnegative reals such that
$$ \sum_{j \ge 0} a_{nj} = \dfrac{1}{n}$$
holds for all naturals $n$?
My progress: I could show that $a_n\le \frac{1}{2n}$. I am not sure if this is even useful.

Let's put
$$
f(z) = \sum\limits_{0\, \le \,j} {\,a_{\,j} z^{\,j} } 
$$
Then the application of the  Series Multisection tells us that
$$
\sum\limits_{0\, \le \,j} {\,a_{\,n\,j} z^{\,n\,j} }  = {1 \over n}\sum\limits_{0\, \le \,k\, \le \,n - 1} {f(\omega ^{\,k} z)} 
$$
Therefore
$$
\sum\limits_{0\, \le \,j} {\,a_{\,n\,j} z^{\,n\,j} }  = {1 \over n}\sum\limits_{0\, \le \,k\, \le \,n - 1} {f(\omega _{\,n} ^{\,k} z)} \quad \left| {\;\omega _{\,n}  = e^{\,i2\pi /n} } \right.
$$
and we are looking for a function $f(z)$ such that
$$
\sum\limits_{0\, \le \,k\, \le \,n - 1} {f(\omega _{\,n} ^{\,k} )}  = 1
$$
which means that the function takes on values over the roots of $1^n$ which sum to $1$ for all values of $n$.
Now, the roots corresponding to $k/n = 1/2, 2/4, 3/6, ...$ have the same value, as all those for which $k/n = const.$.
Consequently also the function will have the same value on those roots.
I am not an expert in the theory of functions, but definitely $f(x)$ cannot be continuous.
