9-3-1. The numbers from 1 to 217 are divided into two groups: one group has 10 numbers, and the other has 207. It turns out that the arithmetic means of the numbers in the two groups are equal. Find the sum of the numbers in the group of 10 numbers.
Answer: 1090.

Solution Variant 1. The shortest solution to this problem is based on the following statement:

If the arithmetic means of the numbers in two groups are equal, then this number is also equal to the arithmetic mean of all the numbers.

A formal proof of this is not difficult: it is sufficient to denote the arithmetic mean by $S$, the quantities of numbers in the groups by $n$ and $k$; then the sums of the numbers in the groups are $S n$ and $S k$, the sum of all the numbers is $S(n+k)$, i.e., their arithmetic mean is again $S$.

However, we urge the reader to understand why this statement is "intuitively obvious." For example, informally, one can think of it this way: in the first group, the numbers are "on average" $S$, and in the second - also. Therefore, if we combine both groups into one, the numbers will be "on average" $S$.

Let's move on to solving the problem. According to our statement, the arithmetic mean in each group is equal to the arithmetic mean of the numbers from 1 to 217, and it is not difficult to check that it is 109. Therefore, the answer is $109 \cdot 10=1090$.