Let the positive integers be A, B, C, and D.  

A + B + C + D = 20

Then there exist positive integers p,q,r and s,
such that

A + B + C = pD
A + B + D = qC
A + C + D = rB
B + C + D = sA

Then we have this system of 5 equations in 8 unknowns:

  A  +  B +  C +  D = 20
  A  +  B +  C - pD =  0
  A  +  B - qC +  D =  0
  A  - rB +  C +  D =  0
-sA  +  B +  C +  D =  0 

Subtracting the 2nd equation from the 1st equation:

D + pD = 20
D(1+p) = 20, so D is a factor of 20

Subtracting the 3rd equation from the 1st equation:

C + qC = 20
C(1+q) = 20, so C is a factor of 20

Subtracting the 4th equation from the 1st equation:

B + rB = 20
B(1+r) = 20, so B is a factor of 20

Subtracting the 5th equation from the 1st equation:

A + sA = 20
A(1+s) = 20, so A is a factor of 20

So all 4 numbers must be factors of 20.

The only factors or 20 are 1,2,4,5,10, and 20.

Studying those for a minute, we see that the only way 
to pick out 4 of them, all different, that have sum 
20, is to leave out the 2 and the 20, and have 

1+4+5+10 = 20.

So the answer is {1,4,5,10}

Checking:

1+4+5 = 10 which is a multiple of 10.
1+4+10 = 15 which is a multiple of 5.
1+5+10 = 16 which is a multiple of 4.
4+5+10 = 19 which is a multiple of 1.

There would be two other answers, if we didn't have to
use all different numbers.  They could all be 5, since
5+5+5+5=20. Also 2+4+4+10=20. 

But since they must all be different, those won't do,
and the only answer is {1,4,5,10} 

Edwin