The answer is $6$. Note that this is possible e.g. with $1,2,4,6,8,10$. On the other hand, if we have $n$ numbers we have $\binom{n}{2}$ pairwise sums. So we certainly need $n \ge 5$ for this to be at least $10$. If $n=5$, then we need all $10$ pairwise sums to be distinct mod $10$. But the sum of all $10$ pairwise sums is even (it's just four times the sum of the numbers) and hence can't equal $1+2+3+\dots+9=45$ modulo $10$.