Problem 2. Pasha wrote the numbers $4,5,6, \ldots, 14$ 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: 91.

Solution. The sum of the numbers from 4 to 14 is 99. If at least one number is erased, the sum of the remaining numbers does not exceed 95. Let's sequentially consider the options:

- if the sum is 95, then Pasha could have erased only the number 4; then the remaining numbers can be divided into five groups with a sum of 19:

$$
14+5=13+6=12+7=11+8=10+9
$$

- if the sum is 94, then Pasha could have erased only the number 5; then the remaining numbers can be divided into two groups with a sum of 47:

$$
14+13+12+8=11+10+9+7+6+4.
$$

- if the sum is 93, then Pasha could have erased only the number 6; then the remaining numbers can be divided into three groups with a sum of 31:

$$
14+13+4=12+11+8=10+9+7+5
$$

- if the sum is 92, then Pasha could have erased only the number 7; then the remaining numbers can be divided into two groups with a sum of 46:

$$
14+13+11+8=12+10+9+6+5+4.
$$

- if Pasha erased the number 8, then the numbers left on the board have a sum of 91: they could be divided either into 7 groups with a sum of 13, or into 13 groups with a sum of 7, or into 91 groups with a sum of 1; but some group will include the number 14, so the sum in this group will be at least 14; therefore, in this case, it is impossible to divide the numbers into groups with the same sum.