We have that $\frac{a_{1} a_{2} ... a_{i-1} a_{i+1} .. a_{n} - 1}{a_{i}}$ is a positive integer. Multiply each of these and expand the numerator, then we have that $\frac{1}{a_1} + ... + \frac{1}{a_n} - \frac{1}{a_{1} a_{2} ... a_{n}}$ is an integer. It is not difficult to prove that there's only finitely many $n$ for which this holds.

qwerty414 wrote: $\frac{1}{a_1} + ... + \frac{1}{a_n} - \frac{1}{a_{1} a_{2} ... a_{n}}$ is an integer. It is not difficult to prove that there's only finitely many $n$ for which this holds. You have to prove that for a fixed $n$, you have finitely many good n-tuples, not that you have good n-tuples for finitely many $n$. However, how do you prove it?

what qwerty414 said isn't true because of this : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=153142&sid=8cea71e77641a2a807a27dc92d599f6b#p153142 and for the problem: as qwerty414 said $ \frac{1}{a_{1}}+...+\frac{1}{a_{n}}-\frac{1}{a_{1}a_{2}... a_{n}} $ must be integer assume the contrary that we have infinitely many $n$-tuples. let $f(a_{1},...,a_{n})=\frac{1}{a_{1}}+...+\frac{1}{a_{n}}-\frac{1}{a_{1}a_{2}... a_{n}}$. for every $n$-tuples $f(a_{1},...,a_{n})< \frac{n}{2}$ since we have infinitely $n$-tuples then infinitely many of them must to be an natural number less than $\frac{n}{2}$ (for example $m$).for one of them like $(a_{1},...,a_{n})$, let $a_{r}$ to be minimum then $\frac{n}{a_{r}}>m$ then $a_{r}<\frac{n}{m}$ thus infinitely of these $n$-tuples have equal minimum member continue this way we'll reach to two equal $n$-tuples this is contradiction so we are done

Quote: what qwerty414 said isn't true I suppose he only had a typo or some problem in expression.