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](https://mathoverflow.net/questions/227713/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 $k \geq m$?



Harmonic sums and elementary number theory
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 $2n + 2$.




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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.




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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.




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Following the comment of Max Alekseyev, [this OEIS sequence](https://oeis.org/A028229) 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.




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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 $(1, 3, 6)$.




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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 $2, 3, 5, 6, 7, 8, 12, 13$.
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