<s>[INST] 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?
 [/INST] 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:
$$1,2,3,4,6,8,9,10,11,39\\
2,3,4,5,7,9,10,11,12,31\\
3,4,5,6,9,10,11,12,13,23\\
4,5,6,7,10,11,12,13,14,15\\
5,6,7,8,10,12,13,14,7$$
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