Let $n_{i} : i \in [6]$ be a set of $6$ odd positive integers.
Let $S = \sum_{\substack{i \ne j \\ i,j = 1}}^{n} n_{i} n_{j}$. What possible values can this sum take?
Here what I have done.
Let $n_{i} = 2k_{i} + 1$. Also, $S = 2 \sum_{i < j} n_{i} n_{j}$. Hence, $S =2 *( 4 \sum_{i < j} k_{i} k_{j} + 2 \sum_{i < j} k_{i} k_{j} + \sum_{i < j} 1) = 8 \sum_{i < j} k_{i} k_{j} + 4 \sum_{i < j} k_{i} k_{j} + 30$.
Doing some manipulation, I got $S = 4((\sum_{i=1}^{6} k_{i})(\sum_{i=1}^{6} k_{i} + 5) - \sum_{i=1}^{6} k_{i}^{2}) + 30$.
Ultimately, I said that $S$ must be of the form $4t + 30$, for some integer $t$. Is this the right answer, or is there somethin I am not considering?
The expression is clearly $(a_1+a_2+\dots + a_6)^2 - (a_1^2 + \dots a_6^2) \equiv 0 - 2 \bmod 4$
Notice that if we change a value by $2$ then the expression changes by $4Z$ where $Z$ is the sum of the other values.
Therefore if we can find a $Z$ such that the possibilities considering only the first $5$ values where $a_1+\dots + a_5 = Z$ include all residue classes $4k \bmod 4Z$ we will have proved that for sufficiently large values all solutions that are $2\bmod 4$ are achievable.
We can take $Z=43$ and it works ( I did the computation). So at least it is true that eventually all can be covered.