Ten natural numbers such that the sums of each choice of nine of them are $82, 83, 84, 85, 87, 89, 90, 91, 92$ In a Whats App group the following puzzle was asked

Ten (not necessarily different) natural number are such that if all but one of them is added, the possible results (depending on which one is omitted) are 82, 83, 84, 85, 87, 89, 90, 91, 92.

Non-uniqueness of the answer was also  mentioned there. Naively, I set up 9 linear equations on 10 variables. By fixing, one of them I could get
the set of 10 numbers as 1,2,3,4,6,8,9,10,11,39. If we leave out first 9 numbers one by one to get the sum of remaining 9  as required. But leaving out 39  does not do so.
What could be the proper approach and the answer?
 A: According to the statement we may assume that the 10 numbers are $$x-5,x-4,x-3,x-2,x,x+2,x+3,x+4,x+5$$
plus a repeated one, say $x+k$, where $k\in\{0,\pm 2,\pm 3,\pm 4,\pm 5\}$. Let $S$ be the sum of those $10$ numbers, then
$$S=(x-5)+(x-4)+\dots+(x+4)+(x+5)+(x+k)=10x+k.$$
Summing all the nine relations together, we obtain
$$S-(x-5)+S-(x-4)+\dots +S-(x+4)+S-(x+5)\\=82+83+84+ 85+87+ 89+90+91+92$$
that is $S-x=(82+83+84+85+87+89+90+91+92)/9=87$, or $10x+k-x=87$,  $9x=87-k$ which implies that $87-k$ should be divisible by $9$. Hence $k=-3$ and therefore $x=10$. It follows that a solution to the puzzle is
$$5,6,7,7,8,10,12,13,14,15.$$
P.S. In Arthur's answer, if we keep subtracting $8$ from $39$ and adding $1$ to the other numbers we get:
$$1,2,3,4,6,8,9,10,11,39\\
2,3,4,5,7,9,10,11,12,31\\
3,4,5,6,9,10,11,12,13,23\\
4,5,6,7,10,11,12,13,14,15\\
5,6,7,8,10,12,13,14,7$$
A: Presumably, one of the numbers 82, 83, 84, 85, 87, 89, 90, 91, 92 is repeated. Add one equation to make such a duplicate. If the resulting solution is non-integer, you chose the wrong number to duplicate. Try another.
Or if you don't like trial-and-error, you could note that by adding all ten equations (including the duplicate), each variable appears with a coefficient of 9, so whatever duplicate value you choose it must be one which makes the sum of all the ten results divisible by 9.
Alternately, you found 1,2,3,4,6,8,9,10,11,39. These work as long as 39 is included, but if 39 isn't included, then the result is too small. You can fix this by "moving" a little value from 39 to the others. Subtract 8 from 39 to get 31, and add 1 to all the others. This will not change the value in any of the sums where 39 (now 31) is included, but will increase the sum where 31 isn't included.
Repeat until done.
A: Let $T$ be the sum of all ten numbers. Let the unknown single number be $x$.
Then $T-x$ is the missing sum of nine, and it will be a repeat of one of the given sums of nine.
If we add up all ten sums of nine numbers (of which we know nine sums, and we know the unknown sum of nine is a repeat of one of the given sums of nine), we will get $9\cdot T$ (since each of the ten numbers is in nine of the sums of nine).
So $82+ 83+ 84+ 85+ 87+ 89+ 90+ 91+ 92+(T-x)=9\cdot T$, a multiple of nine.
Then since $82+ 83+ 84+ 85+ 87+ 89+ 90+ 91+ 92=783$ is a multiple of nine, we must have that $T-x$ (the missing sum of nine) is also a multiple of nine. But $T-x$ is a repeat is one of $82, 83, 84, 85, 87, 89, 90, 91, 92$. So the only possibility is $T-x=90$.
Then $9\cdot T=82+ 83+ 84+ 85+ 87+ 89+ 90+ 91+ 92+90=873$.
Then $T=\frac{873}{9}=97$.
Then the original ten numbers are $97$ minus each of the sums of nine:
$15, 14, 13, 12, 10, 8, 7, 6, 5, 7$
A: Let ten natural numbers be $a,b,c,d,e,f,g,h,k,z$, If $S$ denotes  their}sum then we have $$S-a=82, S-b=83, S-c=84, S-d=85, S-e=87, S-f=89, S-g=90, S-h=91, S-k=92.$$
Adding these 9 equations we $$9S-(S-z)=783\implies S=\frac{783-z}{8}$$
Using this in the 9 equations, we get 9 variables in terms of 10th variab;w $z$ in descending order
$$\frac{127-z}{8},\frac{119-z}{8},\frac{111-z}{8},\frac{103-z}{8},\frac{87-z}{8}, \frac{71-z}{8}, \frac{63-z}{8}, \frac{55-z}{8}, \frac{47-z}{8}.
$$
Let us choose  a natural number $z$ such that  the lowest one attains the  largest value, with all being natural numbers. For $z=7,$
we get ten natural numbers $15,14,13,12,10,8,7,6,5$ and $z=7$ as the required answer. Interestingly, the double entry of 7, occurs by itself.
