Problem :- Are there infinitely many integers $x_1, x_2, x_3,..., x_5$ such that $\sum_{i=1}^{5} \frac{1}{x_i}=\sum_{i=1}^{5} x_i$ and sum of no two terms in $x_i$ is $0$ and $x_i$ is never $1$ or $-1$ . and $(x_1, x_2,. .., x_5)=1$
The original problem was posted here Sum of integers=sum of their reciprocals . However that problem after the pattern was countered by some people in comments, seemed to have reached a dead end, however I was still wondering whether there are infinitely many solutions , and possibly generalize to more than 5 terms .
I just tried to make cases on the number of negative terms as in the previous problem .
Edit: As for the generalization part using n=6 solution mentioned in the linked post[comment section] note and $n=5$ we can compose other solutions for example $\frac{1}{-40k}+\frac{1}{-3k}+\frac{1}{5k}+\frac{1}{8k}+\frac{1}{30k}+\frac{1}{2}+\frac{1}{-4}+\frac{1}{-9}+\frac{1}{-10}+\frac{1}{-15}+\frac{1}{36}$ would be a valid construction for $n=11$ under certain infinitely many values of $k$ to make sure no two terms sum to 0. and as we paste another $\frac{1}{-40a}+\frac{1}{-3a}+\frac{1}{5a}+\frac{1}{8a}+\frac{1}{30a}$ and so $\frac{1}{-40a}+\frac{1}{-3a}+\frac{1}{5a}+\frac{1}{8a}+\frac{1}{30a}+\frac{1}{-40k}+\frac{1}{-3k}+\frac{1}{5k}+\frac{1}{8k}+\frac{1}{30k}+\frac{1}{2}+\frac{1}{-4}+\frac{1}{-9}+\frac{1}{-10}+\frac{1}{-15}+\frac{1}{36}$ would become valid example for $n=16$ and in general we have construction for $n=5k+1$ we can similarly paste the 6 term solution again and again along with the 5 term solution and get a solution for $n=5a+6b$ by Mc Chicken nugget theorom https://artofproblemsolving.com/wiki/index.php/Chicken_McNugget_Theorem . All $n \geq 20$ can be written in that form so we now have infinite solutions for them [note again we have restrictions on constants we are multiplying with to make sure no two sum to $0$[gcd is always obeyed as we can have two coprime terms in the sequence always],but even with that we have infinite].
there are some other ways to compose solutions for $n=31$ using only $n=5$ manually without writing code by which we have examples for $n \geq 120$ using same method.
This is the method we know
$1+\frac{1}{30}+\frac{1}{8}+\frac{1}{5}+\frac{1}{-3}+\frac{1}{-40}=1+30+8+5-3-40=1$ Square both sides without grouping that is for example dont write $a+a=2a$ and keep it as $a$ and $a$ and separetely in the new sequence we have $6^2=36$ terms only two terms that will cancel[sum to 0] will be those in pairs of [observe in the expansion we again have sum of integers=sum of reciprocals] of $8*5$ and $-40*1$ , as we didnt group total 4 terms got removed , and no two other terms sum to 0 , also remove the $1^2$ from both sides , u are left with $36-4-1=31$ terms and gcd is 1 as one as $5^2$ and $8^2$ were in expansion and they are coprime. Hope i was clear coz this part was a bit tough for me to write ask on comments for any quiries.
Imma also share a list of solutions that got produced when i coded for $n=5$[the sum of terms being 0 is not a coincidence but forced to make the search program faster]
some solutions are permutations of others.
Update: Have a proof for n=9 .
Take any tuple $(x, y, z)$ sum of no two terms is 0 of integers none is $1$ or $-1$ .
And $x+y+z=0$ .
Take another such tuple $(a, b, c)$
We show infinite solutions for when the sum is $0$ .
hence it suffices to find infinite rational solutions[which are not just created by multipltying one solution with a constant ] as the equation is homogenous .
Now consider
$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(a+b+c)=(x+y+z)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=0$
When u expand left side u get some sum of rational numbers u have exactly $3^2=9$ terms take the lcm we have something like [rational numbers produced on expansion of RHS are reciprocals of LHS]
$q(\frac{1}{a1}+.... +\frac{1}{a_9})=(a_1+a_2+.... +a_9)\frac{1}{q}=0$
That implies
$\frac{1}{a_1}+... +\frac{1}{a_9}=a_1+a_2+.... +a_9$
We can add further conditions on $(x, y, z)$ and $(a, b, c)$ To ensure no two sum to 0 and gcd=1 and no term is 1 or -1 but those are easy to ensure maintaining infinite solutions .
Previously had posted this as an answer,if i wasent clear enough in it as it got one downvote . Then here is a example to illustrate better
take $(x,y,z)=(3,4,-7)$ and $(a,b,c)=(3,5,-8)$
$0=(3+4-7)(\frac{1}{3}+\frac{1}{5}-\frac{1}{8})=1+3/5-3/8+4/3+4/5-4/8-7/3-7/5+7/8$
and $0=(\frac{1}{3}+\frac{1}{4}-\frac{1}{7})(3+5-8)=1+\frac{5}{3}-\frac{8}{3}+\frac{3}{4}+\frac{5}{4}-\frac{8}{4}-\frac{3}{7}-\frac{5}{7}+\frac{8}{7}$
as $1=120/120$,$3/5=(24*3)/120$ similarly simplify other terms
U get
$\frac{120+24*3-15*3+4*8*5+4*8*3-40*7-4*15-7*24+7*15}{120}=120(1/(120)+1/(24*3)-1/(15*3)+1/(4*8*5)+1/(4*8*3)-1/(40*7)-1/(4*15)-1/(7*24)+1/(7*15))=0$
Automatically giving solution $[120,24*3,-15*3,4*8*5,4*8*3,-40*7,-4*15,-7*24,7*15]$ U can verify no two sum to 0.
More update:-
After some work i showed infinite solutions for $n=7$ and infinite soluitions to the orginal problem but rather the condition that
$(\sum_{i=1}^{5} x_i)/(\sum_{i=1}^{5} \frac{1}{x_i})$ is a non zero perfect square