Answer: 5
Explanation: The modulo w.r.t. 7 of any natural number belongs to one of the 4 sets from ##A=\{\{0\}, \{1, 6\}, \{2, 5\}, \{3, 4\}\}##, i.e. sets of the form ##\{r, (7-r) \mod 7\}## where ##r## belongs to the set of all possible remainders w.r.t. 7, i.e. {0, 1, 2, ..., 6}. Consider 2 distinct natural numbers ##a, b##. It is easy to see that if ##(a \mod 7)## and ##(b\mod 7)## belong to 2 different sets in ##A##, then neither their sum nor difference is divisible by 7. On the other hand, if they belong to the same set from within ##A##, then 2 possibilities arise:
- ##(a \mod 7) = (b \mod 7)##, i.e. the remainders w.r.t. 7 of both ##a## and ##b## correspond the same element in the set. In this case, ##(a - b) \equiv 0 \mod 7##, i.e. the difference is divisible by 7.
- ##(r_{a} \equiv (a \mod 7)) \neq (r_{b} \equiv (b \mod 7))##. Then, by definition of the sets, ##r_{a} = (7 - r_{b}) \mod 7##. Therefore, ##(r_{a} + r_{b}) = 7 \equiv 0 \mod 7 \Rightarrow (a+b) \equiv 0 \mod{7}##, i.e. the sum is divisible by 7.
Thus, for neither sum nor difference of ##a, b## to be divisible by 7, their modulo w.r.t. 7 must belong to 2 different sets from ##A##. By pigeonhole principle, there can be at most 4 distinct natural numbers in a set such that the modulo w.r.t. 7 of no two of those belong to the same set from ##A##. Thus, if we have a set of 5 (or more) natural numbers, at least two of those numbers will have either their sum or their difference divisible by 7.