I think the solution given is of a weaker, rather than stronger, result. Here, I hope, is a proof of the proposition as stated.

Assume the numbers chosen are a<b<c<d<e<f<g

The maximum sum of 6 chosen numbers is 24+23+22+21+20+b=110+b and so the maximum sum which could be repeated is 109+b.

The smallest possible sum which could be repeated is 1+b since this could be equal to c.

Therefore there are at most 109 different totals but the number of subsets under consideration is 2^7-6=122 since each of {}, {a}, {b}, {b,c,d,e,f,g}, {a,c,d,e,f,g} and {a,b,c,d,e,f,g}cannot have the same sum as another set.

So there are 122 pigeons and only 109 pigeonholes.

Why? The problem asks to show there are repetitions. The proof shows that there repetitions even if the consideration is restricted to 4-element sets.

>
>Assume the numbers chosen are a<b<c<d<e<f<g
>
>The maximum sum of 6 chosen numbers is
>24+23+22+21+20+b=110+b and so the maximum sum which could be
>repeated is 109+b. >and 21 in our set but there is some slack so we can afford
>to be a bit wasteful and overlook that.]
>
>The smallest possible sum which could be repeated is 1+b
>since this could be equal to c.
>
>Therefore there are at most 109 different totals but the
>number of subsets under consideration is 2^7-6=122 since
>each of {}, {a}, {b}, {b,c,d,e,f,g}, {a,c,d,e,f,g} and
>{a,b,c,d,e,f,g}cannot have the same sum as another set.
>
>So there are 122 pigeons and only 109 pigeonholes.