Problem: Find all positive integers $n$ for which there exist an integer $k \ge 2$
and positive rational numbers $a_1$, \dots, $a_k$ satisfying
$a_1 + a_2 + \dots + a_k = a_1 a_2 \dots a_k = n$.

Solution: The answer is all $n$ other than $1,2,3,5$.

\begin{claim*}
  The only solution with $n \le 5$ is $n = 4$.
\end{claim*}
\begin{proof}
  The case $n=4$ works since $2+2 = 2\cdot2=4$. So assume $n > 4$.

  We now contend that $k > 2$.
  Indeed, if $a_1 + a_2 = a_1 a_2 = n$ then
  $(a_1-a_2)^2 = (a_1+a_2)^2 - 4a_1a_2 = n^2-4n = (n-2)^2 - 4$
  is a rational integer square, hence a perfect square.
  This happens only when $n = 4$.

  Now by AM-GM,
  \[ \frac nk = \frac{a_1 + \dots + a_k}{k}
    \ge \sqrt[k]{a_1 \dots a_k} = n^{1/k} \]
  and so $n \ge k^{\frac{1}{1 - 1/k}} = k^{\frac{k}{k-1}}$.
  This last quantity is always greater than $5$, since
  \begin{align*}
    3^{3/2} &= 3\sqrt3 > 5 \\
    4^{4/3} &= 4\sqrt[3]{4} > 5 \\
    k^{\frac{k}{k-1}} &> k \ge 5 \qquad \forall k \ge 5.
  \end{align*}
  This finishes the proof.
\end{proof}

Now, in general:
\begin{itemize}
  \ii If $n \ge 6$ is even, we may take
  $(a_1, \dots, a_{n/2}) = (n/2, 2, 1, \dots, 1)$.
  \ii If $n \ge 9$ is odd, we may take
  $(a_1, \dots, a_{(n-3)/2}) = (n/2, 1/2, 4, 1, \dots, 1)$.
  \ii A special case $n = 7$: one example is $(4/3, 7/6, 9/2)$.
  (Another is $(7/6, 4/3, 3/2, 3)$.)
\end{itemize}

\begin{remark*}
  The main hurdle in the problem is the $n=7$ case.
  One good reason to believe a construction exists is that
  it seems quite difficult to prove that $n=7$ fails.
\end{remark*}