Find four distinct integers greater than 0 such that when added in groups of three, they give $6, 7, 8$, and 9.
Solution 1 - Initially observe that if the largest sum of three of these numbers is 9, then all the numbers must be less than 7, that is:

$$
1,2,3,4,5,6
$$

On the other hand, if the smallest sum is 6, then they must be less than 5, so the remaining numbers are:

$$
1,2,3,4
$$

We verify that these are the numbers:

$$
1+2+3=6,1+2+4=7,1+3+4=8,2+3+4=9
$$

Solution 2 - Adding three at a time four numbers \(a, b, c\) and \(d\) we get the numbers \(a+b+c, a+b+d, a+c+d\) and \(b+c+d\). Therefore,

\(6+7+8+9=(a+b+c)+(a+b+d)+(a+c+d)+(b+c+d)=3(a+b+c+d)\).

Thus, \(a+b+c+d=\frac{30}{3}=10\). Therefore, the numbers sought are

$$
10-6=4 \quad ; \quad 10-7=3 \quad ; \quad 10-8=2 \quad ; \quad 10-9=1
$$