The number of ordered pairs $\left(a,b\right)$ of positive integers such that $\frac{2a-1}{b}$ and $\frac{2b-1}{a}$ are both integers is 

For positive integers '$a$' and '$b$' the numbers $\frac{2a-1}{b}$ and $\frac{2b-1}{a}$ are integers if '$a$' and '$b$' and odd integers because '$2a-1$' and '$2b-1$' are odd integers. 

Now, let $\frac{2a-1}{b}=\alpha$ and $\frac{2b-1}{a}=\beta$, where $\alpha ,\beta$ are integers.

then $2a-1=\alpha b$ and $2b-1=\beta a$

so $4a-2=\alpha \left(\beta a+1\right)$

$\Rightarrow a=\frac{\alpha +2}{4-\alpha \beta }$

$\because$ $a$ is an integer, then $0<\alpha \beta <4$

$\therefore$ Possible value of $\alpha =1, 2$ or $3$ and $\beta =1, 2$ or $3$

such that $0<\alpha \beta <4$

Now, when $\left(\alpha ,\beta \right)=\left(1,1\right)$, then $\left(a,b\right)=\left(1,1\right)$

When $\left(\alpha ,\beta \right)=\left(1,2\right)$, then $\left(a,b\right)$ have no value

When $(\alpha ,\beta )=(1,3)$, then $\left(a,b\right)=\left(3,5\right)$ and similarly when $\left(\alpha ,\beta \right)=\left(3,1\right)$, then $\left(a,b\right)=\left(5,3\right)$ 

So, number of ordered pairs $\left(a,b\right)$ is $3$.