Autumn tournament

a) The quadruple $(7,6,4,3)$ is nice because there are no three numbers in it to form an arithmetic progression, but of the six numbers $$7+6=13,7+4=11,7+3=10,6+4=10,6+3=9,4+3=7$$ the numbers $7,9,11,13$ form an arithmetic progression.

b) Without restriction let $a>b>c>d$. Then $$a+b>a+c>\max (a+d,b+c)>\min (a+d,b+c)>b+d>c+d.$$

1) $a+b,a+c$ and $a+d$ form an arithmetic progression, then $2(a+c)=(a+b)+(a+d)⟺2c=b+d$; 2) $a+b,a+c$ and $b+c$ form an arithmetic progression, then $2(a+c)=(a+b)+(b+c)⟺2b=a+c$; 3) $a+d,b+d$ and $c+d$ form an arithmetic progression, then $2(b+d)=(a+d)+(c+d)⟺2b=a+c$; 4) $b+c,b+d$ and $c+d$ form an arithmetic progression, then $2(b+d)=(b+c)+(c+d)⟺2c=b+d$. In all four cases we get a contradiction with the given condition. From the above it is clear that all 6 numbers cannot form an arithmetic progression. Let's assume that 5 of them form an arithmetic progression. Notice that whichever of the numbers $a+b,a+c,a+d,b+c,b+d$ and $c+d$ we delete, there is always some from progressions 1., 2., 3., or 4., contradiction. It follows from a) that the required $k$ is 4. $\square$