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!).

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.

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!).

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!).