For three continuous natural numbers, the sum of the square difference of the larger and smaller numbers and the middle natural number must be the multiple of that number? 1,3,4,5

For three continuous natural numbers, the sum of the square difference of the larger and smaller numbers and the middle natural number must be the multiple of that number? 1,3,4,5

Because the result is a positive integer, it must be a multiple of 1 first
Suppose that the middle number is a (a > 2 natural number)
(a+1)²-(a-1)²+a
=4a+a
=5a
So it must be a multiple of five

The sum of three continuous natural numbers must be a multiple of 3______ (judge right or wrong)

Let the first of the three continuous natural numbers be a, then the sum of the three continuous natural numbers is a + (a + 1) + (a + 2) = 3 × (a + 1). Therefore, the sum of the three continuous natural numbers must be a multiple of 3

Any given 2007 natural numbers. Proof: there must be a number of natural numbers, and is a multiple of 2007 (a single number is also regarded as a sum)
(please specify the reasons)

Let the 2007 numbers be A1, A2,..., a2007 as a sequence, A1, a1 + A2, a1 + A2 + a3,..., a1 +... + a2007, then there are 2007 numbers in the sequence. Then we discuss the classification of 1. If there is a number in the sequence, a1 +... An is a multiple of 2007, then the proposition is right. 2