In a given group of numbers, one number is equal to the average of all, the largest number is 7 greater than the average, the smallest is 7 less than the average, and most of the numbers in the group have below-average values.

What is the smallest number of numbers that can be in the group?

(K. Pazourek)

Hint. What is the average of the three more closely described numbers from the group?
Let's denote the average of the numbers in the group as $p$. The smallest number in the group is $p-7$, and the largest is $p+7$. The average of these three numbers is $p$, so the average of the remaining numbers in the group must also be $p$.

Therefore, some of the remaining numbers must be less than $p$, and some must be greater than $p$. To ensure that the majority of the numbers are below average, there must be at least two more numbers that are less than $p$ than those that are greater than $p$.

The group contains at least seven numbers, arranged schematically as follows:

$$
p-7, \quad *, \quad *, \quad *, \quad p, \quad *, \quad p+7 .
$$

Notes. There are infinitely many sets of seven numbers that satisfy the conditions; they can be described, for example, as follows:

$$
p-7, \quad p-a, \quad p-b, \quad p-c, \quad p, \quad p+d, \quad p+7
$$

where $0 < a, b, c, d \leq 7$ and $a+b+c=d$. (An example could be the set $-7, -4, -1, -1, 0, 6, 7$.)

The formal justification for the initial observation follows from the definition of (arithmetic) average: if the number of remaining numbers in the group is $n$ and their sum is $s$, then their average is $\frac{s}{n}$, while the average of all numbers is $\frac{s+3p}{n+3}=p$. By manipulating the second expression, we get $\frac{s}{n}=p$.

The problem can be solved by incrementally increasing the number of numbers in the group and verifying all the given conditions.