Find four positive integers such that when added in groups of three, they give sums of $6, 7, 8$, and 9.
Adding four numbers $a, b, c$ and $d$ in groups of three, we obtain the numbers $a+b+c, a+b+d, a+c+d$ and $b+c+d$, such that

$$
(a+b+c)+(a+b+d)+(a+c+d)+(b+c+d)=3(a+b+c+d)
$$

Since $30=6+7+8+9=(a+b+c)+(a+b+d)+(a+c+d)+(b+c+d)$, it follows that $30=3(a+b+c+d)$, hence $a+b+c+d=\frac{30}{3}=10$. As each number is equal to the difference between the sum of the four numbers and the sum of the other three, the numbers sought are

$$
10-6=4, \quad 10-7=3, \quad 10-8=2 \quad \text { and } \quad 10-9=1
$$