Q:

When $ab/(a+b)$ is an integer, where $a,b$ are positive integers. 

When $ab/(a+b)$ is an integer, where $a,b$ are positive integers? 
clear;
maxn:=30;
for a in [1..maxn] do
for b in [a..maxn] do
q1:=a*b; q2:=a+b;
if q1 mod q2 eq 0 then
  print a,b,q1 div q2;
end if;
end for;
end for;

the Magma code given above outputs the following. 
2 2 1
3 6 2
4 4 2
4 12 3
5 20 4
6 6 3
6 12 4
6 30 5
8 8 4
8 24 6
9 18 6
10 10 5
10 15 6
12 12 6
12 24 8
14 14 7
15 30 10
16 16 8
18 18 9
20 20 10
20 30 12
21 28 12
22 22 11
24 24 12
26 26 13
28 28 14
30 30 15

So I conjectrue that $\frac{ab}{a+b}$ is an integer if and only if $\frac{a}{d}+\frac{b}{d} \mid d$, where $d=\gcd(a,b)$. It is true if $a=b$.
If $\frac{a}{d}+\frac{b}{d} \mid d$, then 
$$\frac{ab}{a+b}=\frac{a}{d} \cdot \frac{b}{d}\cdot \frac{d}{\frac{a}{d}+\frac{b}{d}}$$
is a product of three positive integers and hence is an integer. 
Conversely, i.e., $\frac{ab}{a+b}$ is an integer.  Let $b=\frac{p}{q}\cdot a$ with $\gcd(p,q)=1$ where $p=\frac{b}{d}$, $q=\frac{a}{d}$ and $d=\gcd(a,b)$. Then
$$\frac{ab}{a+b}=\frac{a\cdot \frac{p}{q}\cdot a }{(1+\frac{p}{q})a}=\frac{ap}{p+q}$$
Since $\gcd(p,p+q)=\gcd(p,q)=1$, it has $p+q \mid a$. Similarly, it has $p+q \mid b$. As a result, $p+q \mid \gcd(a,b)$, i.e. $\frac{a}{d}+\frac{b}{d} \mid d$.
It completes the proof.  

A:

Yes, your conjecture is correct.  Below is a simple proof. 
Theorem $\,\ A\!+\!B\mid AB\iff  a+b\mid c,\  \ c := (A,B),\,\ a,\,b := A/c,\, B/c $ 
Proof $\,\ \dfrac{AB}{A+B}  = \dfrac{acbc}{ac+bc} = \dfrac{abc}{a+b}\ $ is an integer
$\ \iff\, a\!+\!b\mid \color{#0a0}a\color{#c00}bc\iff a\!+\!b\mid c,\, $ by $\,(a\!+\!b,\color{#0a0}a)=\underbrace{(a,b)}_{\large 1} = (a\!+\!b,\color{#c00}b)\,$ by Euclid

