4. 1. 9 * If the system of inequalities $\left\{\begin{array}{l}9 x-a \geqslant 0, \\ 8 x-b<0\end{array}\right.$ has only the integer solutions $1,2,3$, then how many ordered pairs $(a, b)$ of integers satisfy this system of inequalities?
From the original system of inequalities, we can solve to get $\frac{a}{9} \leqslant x<\frac{b}{8}$. Plotting this solution set on a number line, as shown in the figure, we get $0<\frac{a}{9} \leqslant 1, 3<\frac{b}{8} \leqslant 4$, which means $0<a \leqslant 9, 24<b \leqslant 32$. Therefore, $a=1,2, \cdots, 9 ; b=25,26, \cdots, 32$. The ordered pairs $(a, b)$ total $9 \times 8=72$ pairs.