What is the sum of the ages?

Let there be $n$ number of women [where $\color{#D61F06}{n\in\mathbb{N}}$ ]. It is given that the $\text{Total combination} =6$

$\therefore \begin{aligned}& \binom{n}{2}=6 \\& \implies \dfrac{n!}{2!(n-2)!}=6\\& \implies \dfrac{n(n-1)}{2}=6\\& \implies n^2-n-12=0\\& n^2-4n+3n-12=0\\& n(n-4)+3(n-4)=0\\& \implies (n-4)(n+3)=0\\&\\& \color{#3D99F6}{\text{Thus,} ~n=4~\text{or}~-3}\\&\\& \text{Since,}~\color{#D61F06}{n\in\mathbb{N}}:~ \color{#20A900}{n=4}\end{aligned}$

Thus there are $\color{#20A900}{4}$ number of women.

Let thier ages be $a,b,c,d$

$\therefore \text{If} ~a<b<c<d~\begin{cases}& a+b=30 ~~-~~\text{(1)}\\& a+c=33 ~~-~~\text{(2)} \\& a+d=41~~-~~\text{(3)}\\& b+c=58~~-~~\text{(4)}\\& b+d=66~~-~~\text{(5)}\\& c+d=69~~-~~\text{(6)}\end{cases}$

Hence, by $\text{(1)}+\text{(2)}+\text{(3)}+\text{(4)}+\text{(5)}+\text{(6)}$ we have:

$\begin{aligned}&3(a+b+c+d)=297\\& \implies a+b+c+d=\boxed{99}\end{aligned}$