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 $2(a+c) = (a+b) + (a+d) \iff 2c = b+d$;
2) $a+b, a+c$ and $b+c$ form an arithmetic progression, then $2(a+c) = (a+b) + (b+c) \iff 2b = a+c$;
3) $a+d, b+d$ and $c+d$ form an arithmetic progression, then $2(b+d) = (a+d) + (c+d) \iff 2b = a+c$;
4) $b+c, b+d$ and $c+d$ form an arithmetic progression, then $2(b+d) = (b+c) + (c+d) \iff 2c = b+d$.
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. $\square$