Find the largest positive integer which can divide the sum of any five such numbers.

Suppose the positive integers are $a<b<c<d<e$.

Then $2a<a+b<2b<b+c<2c<c+d<2d<d+e<2e$ give nine distinct sums. So these must be the only sums of pairs.

Then for $a+c$ we have $a+b<a+c<b+c$

So $a+c = 2b$

Similar reasoning will show that

$b+d=2c$;

$c+e=2d$;

$a+d=b+c$;

$b+e=c+d$; and

$a+e=2c$.

From all this we obtain: $(c-b)-(b-a)=(a+c-2b)=0$

So $c-b=b-a$.

Similarly we will end up with $e-d=d-c=c-b=b-a$.

So the five numbers are in arithmetic progression. So their sum is divisible by $5$ (the sum will be $5$ times the middle value).

Suppose I have a set of such numbers with sum $S$. If I add $1$ to each of the five numbers, I don't change the fact that the condition is achieved, but I change the sum to $S+5$. This means that the only possible answers to the question are the numbers $1$ and $5$. [This observation is not necessary, but is the kind of useful guide to what might be happening and also check on calculations which comes in handy].

I can spot two chains of nine sums which must be different (one given in another answer). The corresponding terms of the chains must then be equal, and that shows that the terms must be in arithmetic progression. Five consecutive terms of an AP have sum divisible by $5$. I suggest trying to find the second chain for yourself.

Then with $a\gt b\gt c \gt d \gt e$ I have both $$2a\gt a+b\gt a+c\gt a+d\gt a+e\gt b+e\gt c+e\gt d+e \gt 2e$$ and $$2a\gt a+b\gt 2b\gt b+c\gt2c\gt c+d\gt 2d \gt d+e \gt 2e$$ whence $$a+c=2b, a+d=b+c, a+e=2c, b+e=c+d, c+e=2d$$ from which $$a-b=b-c, a-b=c-d, [a-c=c-e], b-c=d-e, c-d=d-e$$ showing that there is a common difference - the middle term is not needed for this.