I suspect that $k = 4$ is good, but am not sure how to prove it. However, every positive integer $k \geq 5$ is good. This follows from the fact (see the proof of Theorem 1 from this preprint) that for any rational number $x$, there are rational numbers $a$, $b$, $c$, $d$ so that $a+b+c+d = 0$ and $abcd = x$. In particular, one can take
$$
a(x) = \frac{2(1-4x)^{2}}{3(1+8x)}, b(x) = \frac{-(1+8x)}{6}, c(x) = \frac{-(1+8x)}{2(1-4x)}, d(x) = \frac{18x}{(1-4x)(1+8x)},
$$
as long as $x \not\in \{1/4, -1/8\}$. (For $x = 1/4$ one can take $(a,b,c,d) = (-1/2,1/2,-1,1)$ and for $x = -1/8$ one can take $(a,b,c,d) = (-2/3,25/12,-1/15,-27/20)$.)

Now, fix $k \geq 5$, let $q \in \mathbb{Q}$ and take $a_{1} = a(q/(k-4))$, $a_{2} = b(q/(k-4))$, $a_{3} = c(q/(k-4))$, $a_{4} = d(q/(k-4))$ and $a_{5} = a_{6} = \cdots = a_{k} = 1$. We have that
$$
a_{1} + a_{2} + a_{3} + a_{4} + \cdots + a_{k} = 0 + a_{5} + \cdots + a_{k} = k-4
$$
and $a_{1} a_{2} a_{3} a_{4} \cdots = \frac{q}{k-4} \cdot 1 \cdot 1 \cdots 1 = \frac{q}{k-4}$. Thus
$$
\left(\prod_{i=1}^{k} a_{i}\right) \left(\sum_{i=1}^{k} a_{i}\right) = \frac{q}{k-4} \cdot (k-4) = q.
$$