For that (a,b)

Here,(1\a) + (1\b) =1\2004
or,a = (b×2004)÷(b+2004)---(i)
As 'a' is a natural number ,
(b+2004)|(b×2004)---(ii) [here,one think to be mentioned - a|b means a divides b] But (b+2004)|2004(b+2004)---(iii)
Now as we know if a|b and a|c , then a|(b-c) . Using this lemma with (ii) and (iii) , we can state that (b+2004)|2004²---(iv) . Now, look, 'b' can be natural if b+2004>-2004 .But ensuring 'b' to be natural 'a' can be natural if (b+2004) is positive . [from (i) , it is evident] Now from (iv) ,each (b+2004) is a positive factor of 2004² . Here , 2004²=2⁴×3²×167² . Means , 2004² has total (4+1)(2+1)(2+1)=45 number of positive factors . So , (a,b) are of 45 types .
Ans:45. 🎃🎃🎃

Multiply both sides by $2004ab$ , then use Simon's favourite factoring trick :

$\frac{1}{a} + \frac{1}{b} = \frac{1}{2004} \Rightarrow 2004b+2004a=ab \Rightarrow ab-2004a-2004b+2004^2 = 2004^2 \Rightarrow (a-2004)(b-2004) = 2004^2.$

There is only one possible value of $b$ for each $a$ . So the question asks for the number of factors of $2004^2$ , and since $2004 = 2^2 \cdot 3 \cdot 167\ \Rightarrow 2004^2 = 2^4 \cdot 3^2 \cdot 167^2$ , the answer is $(4+1)(2+1)(2+1) = \boxed{45}$ .