Problem 2. Dasha wrote the numbers $9,10,11, \ldots, 22$ 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: 203.

Solution. The sum of numbers from 9 to 22 is 217. If at least one number is erased, the sum of the remaining numbers does not exceed 208. Let's sequentially consider the options:

- if the sum is 208, then Dasha could have erased only the number 9; then the remaining numbers can be divided into two groups with a sum of 104:

$$
22+21+20+19+12+10=18+17+16+15+14+13+11
$$

- if the sum is 207, then Dasha could have erased only the number 10; then the remaining numbers can be divided into three groups with a sum of 69:

$$
22+21+17+9=20+19+18+12=16+15+14+13+11
$$

- if the sum is 206, then Dasha could have erased only the number 11; then the remaining numbers can be divided into two groups with a sum of 103:

$$
22+21+20+19+12+9=18+17+16+15+14+13+10
$$

- if the sum is 205, then Dasha could have erased only the number 12; then the remaining numbers can be divided into five groups with a sum of 41:

$$
22+19=21+20=18+13+10=17+15+9=16+14+11
$$

- if the sum is 204, then Dasha could have erased only the number 13; then the remaining numbers can be divided into two groups with a sum of 102:

$$
22+21+20+19+11+9=18+17+16+15+14+12+10
$$

- if Dasha erased the number 14, then the numbers left on the board have a sum of 203: they could be divided into 7 groups with a sum of 29, or 29 groups with a sum of 7, or 203 groups with a sum of 1; one of these groups will contain the number 22; since we have no options with a sum of 22, at least one more number will be added to this group; therefore, the sum in this group will be at least 31; thus, in this case, it is impossible to divide the numbers into groups with the same sum.