, 2015

The integers greater than $3$ are ruled out by noticing that if $a_1, a_2, a_3$ are pairwise distinct positive integers, then
$$\begin{aligned} \frac{(a_1+1)(a_2+1)(a_3+1)-1}{a_1 a_2 a_3} &= 1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \frac{1}{a_1 a_2} + \frac{1}{a_1 a_3} + \frac{1}{a_2 a_3} \\ &\le 1 + 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 3} + \frac{1}{2 \cdot 3} = 3 + \frac{5}{6} \end{aligned}$$
To complete the proof, fix an integer $n \ge 3$, consider an integer $a \ge 3$, and let $a_1 = a - 2$, let $a_k = a^{2^{k-2}}$, $k = 2, \dots, n-1$, and let $a_n = a^{2^{n-2}} - 2$. Then $1 \le a_1 < a_2 < \dots < a_n$, and
$$\begin{aligned} \frac{(a_1+1)(a_2+1)\cdots(a_n+1)-1}{a_1 a_2 \cdots a_n} &= \frac{((a-1) \prod_{k=2}^{n-1} (a^{2^{k-2}}+1)) (a^{2^{n-2}}-1)-1}{(a-2) a^{1+2+\cdots+2^{n-3}} (a^{2^{n-2}}-2)} \\ &= \frac{\left(a^{2^{n-2}}-1\right)^2-1}{(a-2) a^{2^{n-2}-1} (a^{2^{n-2}}-2)} = \frac{a^{2^{n-2}} (a^{2^{n-2}}-2)}{(a-2) a^{2^{n-2}-1} (a^{2^{n-2}}-2)} \\ &= \frac{a}{a-2} = 1 + \frac{2}{a-2}, \end{aligned}$$
which is integral if and only if $a = 3$ or $a = 4$. The former shows that $3$ satisfies the required condition, and the latter shows that so does $2$.