5 days, the date is 2 1 ~ 25 days;
10 day, the date is 7 ~ 16 day;
10 day, the date is 28th to 30th of the following month and1~ 7th;
5 days, the date is 27 ~ 30, and the next month is 1.
First of all, consider the case of not crossing the month, and the date is a continuous positive integer. Let the sum of the dates of the first day and the last day be s, and there will always be * * * D days, and d*s/2 = 1 15, that is, s*d = 230. Calculation shows that only d = 5 or 10 is possible, corresponding to the first two answers.
If it is a cross-month, the situation will be a little more complicated, but it is impossible to include the five days at the end of last month. Otherwise, even in February, it will be at least 24+25+26+27+28= 130, exceeding 1 15. Therefore, the classification discussion includes 1, 2, 3 and 4 days at the end of last month:
If four days at the end of last month are included, the possible sums of these four days are 106, 1 10, 1 4 or 1 18, leaving1 The latter two can't be expressed as the sum of consecutive positive integers starting from 1, so only 27+28+29+30+ 1 can;
If three days at the end of last month are included, the possible sum of these three days is 8 1, 84, 87 or 90, leaving 25, 28, 3 1 or 34. Only 28 can be expressed as the sum of consecutive positive integers starting from 1, so 28+29+30+ 1+2+3+4+5+6+7 can;
If it is two days including the end of last month, the possible sum of these two days is 55, 57, 59 or 6 1, and the remaining 54, 56, 58 or 60 cannot be expressed as the sum of continuous positive integers starting from 1, so there is no solution;
If it is the last day including the end of last month, it is 28, 29, 30 or 3 1, leaving 84, 85, 86 or 87, which cannot be expressed as the sum of consecutive positive integers starting from 1, and there is no solution.
So four possible answers are listed at the beginning.