Could the sum of an even number of distinct odd numbers be divisible by each of the odd numbers ?

Let k≥4 be an even number. Can one find k distinct positive odd numbers x1,…,xk such that each xi divides S=∑ki=1xi ?

Is it possible at least for k big enough ?

**Answer**

Yes, it is possible. Divide your sum by S and you have

1=∑ixiS

which is an Egyptian fraction expression of 1 where all the denominators have the same number of factors of 2. This is known to be solvable with all denominators odd, but all known solutions have an odd number of terms. A survey paper is here. The section of interest is 9.5. One example is:

1=13+15+17+19+111+115+135+145+1231

where the denominators have least common multiple 3465 so we can write:

3465=1155+693+495+385+315+231+99+77+15

with every term dividing the sum. Now if we add 3465 to each side we have a solution with an even number of terms:

6930=3465+1155+693+495+385+315+231+99+77+15

Any Egyptian fraction decomposition of 1 into fractions with odd denominators yields a solution to your problem. The sum will be twice the least common multiple of the denominators in the decomposition. The paper shows that there is such a decomposition for all odd numbers of terms 9 or above. You can multiply any solution by any odd number to get another.

What is happening is we are converting the Egyptian fraction decomposition of 1 with all denominators odd into one that looks like 1=12+12(all other terms)

**Attribution***Source : Link , Question Author : Olórin , Answer Author : Ross Millikan*