Let’s divide the first 100 natural numbers in five sets of 20 numbers each: {1, 2, 3....20}, {21, 22, 23....40}.................{81, 82, 83.......100}. If we pick the first ten numbers from each set we will not get any pair of two numbers whose difference is 10. However, if we pick just one more number from any of the sets, it would have a difference of 10 with one of the numbers which has already been picked. So the answer is 10×5+1=51.