x = n+[(n ^ 2)/(y-n)]

To make x and y positive integers, it is necessary and only necessary that n 2 can be divisible by y-n

That is, y-n is a factor of n 2 (including 1 and n 2 itself).

So each factor of n 2 corresponds to a y, and each y corresponds to an x,

And the number of solutions s(n)=2007, so we know that n 2 has only 2007 factors.

Let the standard decomposition of n be: n = (p1a1) × (p2a2 )×1... × (pnan).

Where pi is a different prime number and ai is the exponent of pi.

Then n 2 = (p12a1) × (p22a2 )× ... × (pn 2an).

Therefore, there is a factor of (2a1+1) × (2a2+1)× n 2.

From the meaning of the title, (2a1+1) ×××× (2an+1) = 2007.

Because there are only four decomposition forms in 2007, namely:

2007= 1×2007

2007=3×3×223

2007=9×223

2007=3×669

So the solution is:

1、a 1= 1003

2、a 1= 1，a2= 1，a3= 1 1 1

3、a 1=4，a2= 1 1 1

4、a 1= 1，a2=334

Therefore, it is found that all positive integers n that make S (n) = 2007 can be expressed in the following four forms:

1、n=p 1^ 1003

2、n=p 1×p2×(p3^ 1 1 1)

3、n=(p 1^4)×(p2^ 1 1 1)

4、n=p 1×(p2^334)