Note that
$$b_n=\dfrac{a_n+|a_n|}{2} \;\;(\geqslant 0), \\
c_n=\dfrac{a_n-|a_n|}{2} \;\;(\leqslant 0)
$$
and
$$c_n=a_n-b_n.$$
If $\sum a_n$ converges (conditionally) and $\sum b_n$ is convergent (absolutely) then $\sum {c_n}$ is convergent (absolutely). Because $a_n=b_n+c_n$ then $\sum a_n$ must be absolutely convergent, which contradicts to its conditional convergence.
Simply having infinitely many non-zero terms is insufficient. There are many cases of infinite series with infinitely many non-zero terms that converge, e.g. $\displaystyle \sum_{n = 1}^\infty \frac{1}{n^p}$ for $p > 1$.
When you say that $\displaystyle \sum_{n = 1}^\infty a_n$ converges conditionally, I presume you mean that it converges but it does not converge absolutely. (As an aside, this already means that we have infinitely many non-zero terms - if there were only finitely many, it would converge absolutely, and thus be an absolutely convergent series).
Let's suppose that $\displaystyle \sum_{n= 1}^\infty a_n^-$ converges for a moment, so that we can say that
$$\sum_{n= 1}^\infty a_n^- = -L$$
for some number $-L$. We know that $\displaystyle \sum a_n$ diverges, so we must have that $\displaystyle \sum a_n^+$ diverges. But if $\displaystyle \sum a_n^+$ diverges, then $\displaystyle \sum a_n = \sum a_n^+ + \sum a_n^-$ will look like $\displaystyle \left( \sum a_n^+\right) - L$, and will still diverge. (Note that I'm being a bit abusive and leaving out details like that these are partial sums; you cannot change the order of elements in a conditionally convergent sum and expect nothing to change).
This contradicts our initial condition that $\displaystyle \sum a_n$ converges. So we must actually have that $\displaystyle \sum a_n^-$ was divergent. $\diamondsuit$
To boil down the essence of the proof, an absolutely convergent series means that all the terms get sufficiently small sufficiently fast to converge. Having a series be merely conditionally convergent means that the positive part and the negative parts cancel out a lot, and without that cancellation you get divergence. By 'cancel out a lot,' I really mean infinite cancellation since this sort of work does not worry about finite contribution. So if either the positive or negative contribution is finite, there isn't enough cancellation.
Best Answer
If there are finitely many positive terms, let's assume the last one is $a_k$. Then we can split the series into $\sum_{n=0}^k a_n$ + $\sum_{n=k+1}^\infty a_n$. Can you show that this second sum, which only has negative terms, converges absolutely?