The answer is $n = 4$ or $n \ge 6$.

I. First, we prove that each $n \in \{4, 6, 7, 8, 9, \dots\}$ satisfies the condition.

(1). If $n = 2k \ge 4$ is even, we set $(a_1, a_2, \dots, a_k) = (k, 2, 1, \dots, 1)$:
$$a_1 + a_2 + \dots + a_k = k + 2 + 1 \cdot (k-2) = 2k = n,$$
and
$$a_1 \cdot a_2 \cdot \dots \cdot a_k = 2k = n.$$

(2). If $n = 2k + 3 \ge 9$ is odd, we set $(a_1, a_2, \dots, a_k) = (k + \frac{3}{2}, \frac{1}{2}, 4, 1, \dots, 1)$:
$$a_1 + a_2 + \dots + a_k = k + \frac{3}{2} + \frac{1}{2} + 4 + (k-3) = 2k + 3 = n,$$
and
$$a_1 \cdot a_2 \cdot \dots \cdot a_k = \left(k + \frac{3}{2}\right) \cdot \frac{1}{2} \cdot 4 = 2k + 3 = n.$$

(3). A very special case is $n = 7$, in which we set $(a_1, a_2, a_3) = (\frac{4}{3}, \frac{7}{6}, \frac{9}{2})$. It is also easy to check that
$$a_1 + a_2 + a_3 = a_1 \cdot a_2 \cdot a_3 = 7 = n.$$

II. Second, we prove by contradiction that each $n \in \{1, 2, 3, 5\}$ fails to satisfy the condition.
Suppose, on the contrary, that there is a set of $k \ge 2$ positive rational numbers whose sum and product are both $n \in \{1, 2, 3, 5\}$. By the Arithmetic-Geometric Mean inequality, we have
$$n^{1/k} = \sqrt[k]{a_1 \cdot a_2 \cdot \dots \cdot a_k} \le \frac{a_1 + a_2 + \dots + a_k}{k} = \frac{n}{k},$$
which gives
$$n \ge k^{\frac{k}{k-1}} = k^{1+\frac{1}{k-1}}.$$
Note that $n > 5$ whenever $k = 3, 4$, or $k \ge 5$:
$$k = 3 \Rightarrow n \ge 3\sqrt{3} = 5.196\dots > 5;$$
$$k = 4 \Rightarrow n \ge 4\sqrt[3]{4} = 6.349\dots > 5;$$
$$k \ge 5 \Rightarrow n \ge 5^{1+\frac{1}{k-1}} > 5.$$
This proves that none of the integers 1, 2, 3, or 5 can be represented as the sum and, at the same time, as the product of three or more positive numbers $a_1, a_2, \dots, a_k$, rational or irrational.
The remaining case $k = 2$ also goes to a contradiction. Indeed, $a_1 + a_2 = a_1a_2 = n$ implies that $n = a_1^2/(a_1 - 1)$ and thus $a_1$ satisfies the quadratic
$$a_1^2 - n a_1 + n = 0.$$
Since $a_1$ is supposed to be rational, the discriminant $n^2 - 4n$ must be a perfect square. However, it can be easily checked that this is not the case for any $n \in \{1, 2, 3, 5\}$. This completes the proof.