Harmonic sums and elementary number theory Which natural number can be represented as a product of a sum of natural numbers and a sum of their inverses? I. e. does there exist for a natural $n$ a set of natural numbers $\{a_1, a_2,...a_m\}$ such that $n = (a_1 + a_2 + ...+a_m)(\frac{1}{a_1} + \frac{1}{a_2} + ... +\frac{1}{a_m})$? Call $n$ good if such a set exists, they do not have to be distinct. 
All $n = k^2$ are good, if n is good then $2n + 2$ is good as well, 10 and 11 are good, 2,3,5,6,7,8 are not. I'm too lazy to check any further especially if there is a solution out there, please point me if you know one. 
Does there exist a constant $C$ such that all $n \geq C$ are good? What if I let $a_i$s to be negative?
Now mathoverflow gave me a link to another question 
Estimating the size of solutions of a diophantine equation which gives 14 as a good number (hard). Moreover there is a link to a very nice paper "An Unusual Cubic Representation Problem" by Andrew Bremner and Allan MacLeod which gives a whole bunch of solutions already for $m = 3$ on page 38 here http://ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf, for all their $N$ we get $2(N + 3)$ to be good. So if we take $C = 30$, then all even $N \geq C$ are good (needs a clear proof). 
Question
There is still question about what happens for odd $N$s. And maybe it's worthy to ask if we can restrict $m$. I. e. What is the smallest $k$ such that any good $N$ can be represented as a product above with $k \geq m$?
 A: You can find an infinite number of good numbers $n$. You can use the formula $$\frac{1}{\frac{2!}{1}}+\frac{1}{\frac{3!}{2}}+\frac{1}{\frac{4!}{3}}+...+\frac{1}{\frac{m!}{m-1}}+\frac{1}{m!}=1$$ and deduce that $n=(2!/1+3!/2+...+m!/(m-1)+m!)(\frac{1}{\frac{2!}{1}}+\frac{1}{\frac{3!}{2}}+\frac{1}{\frac{4!}{3}}+...+\frac{1}{\frac{m!}{m-1}}+\frac{1}{m!})$ is obviously good.
A: Any $n$ which is of the form $4P$ where $P$ is a perfect number is good: just consider as a set $\{a_{i}\}$ the set of its divisors. As the sum of the reciprocals thereof equals $2$, this guarantees that your product is an integer, hence that $n$ is good.
A: Graham (1963) proved that any integer $n>77$ can be represented as
$$ n = a_1 + \dots + a_m,$$ 
where $a_i$ are distinct positive integers with 
$$\frac{1}{a_1} + \dots + \frac{1}{a_m} = 1.$$
Hence, any integer $n>77$ is good.
Btw, I've recently extended the result of Graham to sums of squares.

UPDATE. As for the largest non-good (bad?) number, we know that $8$ is not good, while OEIS A028229 leaves just seven more candidates for the largest non-good number: $12, 13, 14, 15, 19, 21$, or $23$.
Numbers $14, 15, 19, 21, 23$ are good with the corresponding $\{ a_i\}$ being $\{ 2, 3, 10\}$, $\{1, 2, 6\}$, $\{ 5, 8, 12, 15\}$, $\{ 8, 14, 15, 35\}$, and $\{76, 220, 285, 385\}$, respectively.
Hence, there remains just 3 candidate for the largest non-good number: $8, 12$, or $13$. 
While the latter two candidates can be good only with $m=3$, MacLeod's approach (Sec. 6.2) here leads to elliptic curves of zero rank, thus implying that both $12$ and $13$ are not good. 
Therefore, $13$ is the largest non-good number.
A: A note on the observation "$n$ good implies $2n + 2$ good":
First remark is that $n$ is good iff there are positive rational numbers $a_1, \dotsc, a_m$ such that $n = (a_1 + \dotsc + a_m)(1/a_1 + \dotsc + 1 / a_m)$. This is because one can multiply all $a_i$ by the lcm of their denominators.
Second remark is that $n$ is good iff there are positive rational numbers such that $n = 1/a_1 + \dotsc + 1/a_m$ and $a_1 + \dotsc + a_m = 1$. This is because one can multiply all $a_i$ by the inverse of their sum.
Finally, if $n = 1/a_1 + \dotsc + 1/a_m$ with $a_1 + \dotsc + a_m = 1$, then taking $a_i' = a_i / 2$ for $1\leq i \leq m$ and $a_{m + 1}' = 1/2$ gives us $2n + 2$.

Looking at the proof by Graham cited above, I see that this observation is indeed halfway to the actual proof. Namely we need a second result that $n$ good implies $2n + 179$ good, to take care of the odd integers. Then the result follows from the fact that all integers from $78$ to $333$ are good.

Since we don't require the $a_i$'s to be different, we may take $a_i' = a_i / 2$ for $1 \leq i \leq m$, $a_{m + 1}' = 1/3$, $a_{m + 2}' = 1/6$, and get $2n + 9$ good.
This also suggests that we might improve significantly the bound of Graham.

Following the comment of Max Alekseyev, this OEIS sequence says that all positive integers except $2, 3, 5, 6, 7, 8, 12, 13, 14, 15, 19, 21, 23$ are Egyptian, and hence good.
Since our definition of "good" is weaker than "Egyptian", it doesn't a priori exclude the possibility that some of the above listed numbers are still good.

Therefore it only remains to see whether the above numbers are good.
The task is separated into $m = 2, 3, 4$.
For $m = 2$, it is clear that only for $n = 4$, the equation $(x + y)(1/x + 1/y) = n$ has rational solution $(x, y)$.
For $m = 3$, we are led to the equation $(x + y + z)(xy + yz + zx) = nxyz$. Assuming $n > 9$ and choosing the point $[x, y, z] = [-1, 1, 0]$ as the point at infinity, we get an elliptic curve.
A Weierstrass equation is given by $Y^2 = X^3 + (n^2 - 6n - 3)X^2 + 16nX$, where $X = \frac{4n(x + y)}{x + y - (n - 1)z}$, $Y = \frac{4n(n - 1)(x - y)}{x + y - (n - 1)z}$. To get back $[x, y, z]$ from $X, Y$, we have the formula $[x, y, z] = [(n - 1)X + Y, (n - 1)X - Y, 2X - 8n]$.
This curve has six torsion points, which correspond to useless solutions to our equation. Moreover, the Mordell-Weil group has rank $0$ for all the above $n$, except $n = 14, 15$. This proves that $n = 12, 13$ are not good.
For $n = 14$, we have a solution $(3, 10, 15)$.
For $n = 15$, we have a solution $(1, 3, 6)$.

Thanks again to Max Alekseyev, $n = 19, 21, 23$ are all good, with solutions $(5,8,12,15), (8,14,15,35), (76,220,285,385)$, respectively.

Conclusion: A positive integer is good if and only if it is not one of $2, 3, 5, 6, 7, 8, 12, 13$.

