All you need too do in this case is prove that the LCD (Least Common Denominator) of two consecutive integers is 1. It then follows that m consecutive integers have at least m - 1 pairs of integers that have no common divisors greater than 1. But, since I am a Sadistic Mathematician, I shall leave the proof as an exercise for the reader.
Define the that sets can you do the following:
Remove 1 arbritary number from the set then divide the set in to 2 subsets of 5 numbers each and make sure that the sum of the numbers of the first subset equals the sum of the numbers of the second subset.