When x>y, x-y≥xy/ 19.

That is1/y-1/x ≥119.

Then y < 19

We get a set of unequal positive integers from small to large: 1, 2, 3, 4, 5, 6, 7, 8, 9,1,1,12,1.

Then the minimum difference of their reciprocal in the sense of the problem is118-1/a.

Let's verify them one by one.

Obviously,1/3-1/4 =112 ≥119.

And1/4-1/5 =1/20 ≤119,1/3-1/5 = 2//kloc-.

So we got two groups: 1, 2, 3, 4 and 1, 2, 3, 5.

Select the first group 1, 2, 3 and 4. Because n is as large as possible, and there is a tolerance, positive integers satisfying the previous conditions must be as close as possible, that is, the maximum number in this group should be as small as possible.

Then the next smallest positive integer is 6,1/4-1/6 =112 ≥119.

So there are 1, 2, 3, 4, 6.

The next smallest positive integer is 9,1/6-1/9 =118 ≥19.

Therefore 1, 2, 3, 4, 6, 9.

The next smallest positive integer is 19,1/9-119 ≥119.

So 1, 2, 3, 4, 6, 9, 19.

Stop, because if there is a larger positive integer x, then y= 19 contradicts Y < 19.

At this time n=7.