How many sets of at least two elements are there, consisting of consecutive positive integers, where the sum of the elements is 100?
Solution. If the sum of consecutive numbers is even, then there must be an even number of odd numbers among them. We need to examine two cases further.

a) We add an even number of consecutive numbers. In this case, the stated condition is only satisfied if the number of numbers is divisible by 4 (i.e., we add 4, 8, 12, ... numbers, so among them there are 2, 4, 6, ... odd numbers). The sum of four consecutive numbers is $a + (a + 1) + (a + 2) + (a + 3) = 4a + 6$. This cannot be 100 because it is not divisible by 4. The sum of eight consecutive numbers, similarly calculated, is $8a + 28$, if this is 100, then $a = 9$; the set of numbers is $\{9, 10, 11, 12, 13, 14, 15, 16\}$. The sum of twelve consecutive numbers, similarly calculated, is $12a + 66$. This cannot be 100 because it is not divisible by 4. The sum of 16 or more consecutive numbers cannot be 100 because even the sum of the 16 smallest positive consecutive integers is already more than 100.

b) We add an odd number of consecutive numbers. In this case, the sum of the numbers is a multiple of the middle number, as many times as there are numbers. If the sum of the numbers is 100, then the middle number must be a multiple of 100, and an odd multiple at that. Thus, the possible values for the middle number are: 4 and 20. In the case of 4, there would be 25 numbers, but among them there would also be negatives, so this does not provide a solution. In the other case, we get the set of numbers $\{18, 19, 20, 21, 22\}$. Therefore, there are two sets of numbers that meet the conditions.