Problem 2. Sasha wrote the numbers $7, 8, 9, \ldots, 17$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remaining numbers on the board?
Answer: 121.

Solution. The sum of numbers from 7 to 17 is 132. If at least one number is erased, the sum of the remaining numbers does not exceed 125. Let's sequentially consider the options:

- if the sum is 125, then Sasha could have erased only the number 7; then the remaining numbers can be divided into five groups with a sum of 25:

$$
8+17=9+16=10+15=11+14=12+13
$$

- if the sum is 124, then Sasha could have erased only the number 8; then the remaining numbers can be divided into two groups with a sum of 62:

$$
17+16+15+14=13+12+11+10+9+7
$$

- if the sum is 123, then Sasha could have erased only the number 9; then the remaining numbers can be divided into three groups with a sum of 41:

$$
17+16+8=15+14+12=13+11+10+7
$$

- if the sum is 122, then Sasha could have erased only the number 10; then the remaining numbers can be divided into two groups with a sum of 61:

$$
17+16+15+13=14+12+11+9+8+7
$$

- if Sasha erased the number 11, then the numbers left on the board have a sum of 121: they could be divided either into 11 groups with a sum of 11, or into 121 groups with a sum of 1; but some group will include the number 17, so the sum in this group will be at least 17; therefore, in this case, it is impossible to divide the numbers into groups with the same sum.