Which two-digit positive integers satisfy the condition that the number is six times the sum of the digits of the number that is seven more than it?
Let's find the numbers that satisfy the condition in the form of $10 x+y$, where $1 \leq x \leq 9$ and $0 \leq y \leq 9$, integers. To determine the sum of the digits of the number that is 7 greater than it, it is useful to examine three cases.

1. case: Adding seven does not change the tens digit, i.e., $y \leq 2$. In this case, according to the condition,

$$
10 x+y=6 \cdot(x+y+7), \quad 4 x=5 y+42
$$

However, $x \leq 9, y \geq 0$ means this cannot be true.

2. case: Adding seven changes the tens digit, i.e., $y \geq 3$, but the result remains a two-digit number. In this case,

$$
10 x+y=6 \cdot(x+1+y+7-10), \quad 4 x=5 y-12
$$

It is clear that $4 \mid y$, so with $y \geq 3$, $y$ can only be 4 or 8. From this, $x_{1}=\frac{5 \cdot 4-12}{4}=2, x_{2}=\frac{5 \cdot 8-12}{4}=7$.

Thus, we have found two numbers: 24 and 78, and they do satisfy the requirements of the problem.

3. case: $10 x+y+7 \geq 100$, i.e., $x=9, y \geq 3$. In this case, from the condition,

$$
90+y=6 \cdot(1+0+y+7-10)
$$

which means

$$
102=5 y
$$

which is impossible.