Let the positive, unequal integers be $a_1 < a_2 < cdots < a_k$ with $a_1+a_2+cdots+a_k = a_1a_2cdots a_k = n$. Obviously if $k = 1$ all positive integers work; suppose from now on that $k > 1$. Note that $a_k ge k$. Then we have $a_1 + a_2 + cdots + a_k < ka_k$, while $a_1a_2cdots a_k ge (k-1)!a_k$. So we must have $ka_k > (k-1)!a_k$, which means that $k > (k-1)!$. This means $k le 3$. If $k = 2$, then we have $a_1 + a_2 = a_1a_2$, or $frac{1}{a_1}+frac{1}{a_2} = 1$. This can easily be seen to have only the solution $(2,2)$, which doesn't satisfy our hypothesis that the $a_i$ be unequal.
The case $k = 3$ is then the only tricky case. If $a_1 > 1$, then we have $a_1+a_2+a_3 < 3a_3$ and $a_1a_2a_3 ge 6a_3$. So $a_1 = 1$. We then must find solutions to $1 + a_2 + a_3 = a_2a_3$. This can be rewritten as $(a_2-1)(a_3-1) = 2$, which as $a_2 < a_3$ are integers immediately gives $a_2 = 2, a_3 = 3$.
In summary, all possible solutions are ${n}$ for all positive integers $n$ and ${1,2,3}$.
History: this is definitely a classic problem; see for example 2006 USA Mathematical Olympiad problem #4 (pdf, problem is on second page), which is your problem with several restrictions removed.
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