Some triplets of distinct positive integers (k,l,m) exist

such that each one of the three numbers divides the sum of the other two.

Define such sets.

Let the integers be (c, b, a) where a > b > c

(b+c)/a < 2 , so we know that (b+c)/a = 1.

Solving for c gives c = a-b.

And b must be greater than a/2, because it is greater than c.

So the integers are (a-b, b, a).

Then (a+c)/b = (2a-b)/b is an integer.

Let (2a-b)/b = k.

Solving for b gives b = 2a/(k+1).

By inspection, the only k for which b is less than a and greater than a/2 is k = 2.

So b = 2a/3 and c = a-b = a/3.

Checking, this works because (a+b)/c = 5.

So any multiple of (1,2,3) works and there are no other solutions (except for permutations).