.
ANSWER. The sum must be multiple of 12.
Proof
Let n be the "central" number in the sequence of 6 consecutive odd numbers
(n-5), (n-3), (n-1), (n+1), (n+3), (n+5).
Then the sum is 6n, and since n is an even number n = 2m, 6n = 12m is a multiple of 12.
Since m can be any prime number, it shows that there is no other common divisor of the sums
of this kind.
Solved.