A plausible positivity After getting stuck with the
previous positivity
(it probably sounds too complex),
I would like to give a version of the problem which is of most interest to me.
Consider a sequence of real numbers
$a_1,a_2,\dots,a_n,\dots$ with absolute values bounded above by the first term $a_1=a>0$,
which satisfies, for all $n=1,2,\dots$,
$$
|A_n|\le A \qquad\text{where}\quad A_n=a_1+a_2+\dots+a_n.
$$
In addition, assume that infinitely many terms of the sequence are nonzero.
These settings and Dirichlet's convergence test
guarantee that the series
$$
\sum_{n=1}^\infty\frac{a_n}n
$$
converges.
Assume, in addition, that
$$
\max_{1\le k\le n}A_k+\min_{1\le k\le n}A_k\ge0
\qquad\text{for all}\quad n=1,2,\dots.
$$
The problem is to show that
$$
\sum_{n=1}^\infty\frac{a_n}n>0
$$
and to provide, in terms of $a$ and $A$, a lower (strictly positive) bound for the series.
(The latter is optional, as I am not sure that such a bound exists.)
 A: The sum $\sum a_n/n$ can be negative. Below I construct a finite sequence; one can always add a negligibly small tail to get infinitely many non-zeroes.
Begin with $a_1=1$ and $a_2=-1$.
This gives $A_2=0$ and the partial sum of the main series is $1-1/2=1/2$.
Then, repeat 100 times the following procedure:
Pick an integer $k$ larger than the length of the sequence so far.
Extend $(a_n)$ by zeroes up to $n=10k-1$.
Then set $a_n=1$ for all $n$ from $10k$ to $11k-1$ and $a_n=-1$ for all $n$ from $11k$ to $13k-1$. The $k$ ones contribute less than $1/10k$ each to the main series $\sum a_n/n$, and this is less than $1/10$ in total. The $2k$ negative ones contribute absolute value at least $1/13k$ each, this sums up to at least $2/13$ of negative amount. So the partial sum of the main series went down by at least $2/13-1/10>1/20$.
But we have $A_n=-k$ now (for $n=13k-1$). To fix this, extend $(a_n)$ by a huge amount of zeroes, followed by $k$ ones, so that the contribution of these ones to the main sum is less than $1/100$. 
Now we extended the sequence so that the last $A_n$ is zero again but the partial sum of the main series went down by at least $1/30$. Choose the next $k$ and repeat (finitely many times!).
[Edit] Since the sequence is finite, one can define $A=\max|A_n|$ to satisfy the condition $|A_n|\le A$. The $\max+\min$ condition is immediate from the construction.
