We have ab = 20(a + b) = 20a + 20b,.......(*)
so
ab - 20a = 20b,
a(b - 20) = 20 b,
a = 20b/(b - 20).
a is to be positive so b must be greater than 20, and since the roles of a and b can be reversed a must also be greater than 20.
Let a = 20 + m, and b = 20 + n, then, substituting into (*),
(20 + m)(20 + n) = 20(20 + m + 20 + n),
400 + 20m + 20n + mn = 800 + 20m + 20n.
mn = 400.
So we are looking for two positive integers having a product of 400.
(1) m = 1, n = 400 from which a = 21, b = 420.
(2) m = 2, n = 200 ............... a = 22, b = 220.
(3) m = 4, n = 100 ............... a = 24, b = 120.
etc.
There are five others, (fifteen in all, counting reversals).