Solution:
We have
$$\frac{77}{17}<\frac{\beta}{\alpha}<\frac{197}{43}$$
That is,
$$4+\frac{9}{17}<\frac{\beta}{\alpha}<4+\frac{25}{43}$$
Thus $4<\frac{\beta}{\alpha}<5$. Since $\alpha$ and $\beta$ are positive integers, we may write $\beta=4 \alpha+x$, where $0<x<\alpha$. Now we get
$$4+\frac{9}{17}<4+\frac{x}{\alpha}<4+\frac{25}{43}$$
So $\frac{9}{17}<\frac{x}{\alpha}<\frac{25}{43}$; that is, $\frac{43 x}{25}<\alpha<\frac{17 x}{9}$.
We find the smallest value of $x$ for which $\alpha$ becomes a well-defined integer. For $x=1,2,3$ the bounds of $\alpha$ are respectively $\left(1 \frac{18}{25}, 1 \frac{8}{9}\right),\left(3 \frac{11}{25}, 3 \frac{7}{9}\right),\left(5 \frac{4}{9}, 5 \frac{2}{3}\right)$. None of these pairs contain an integer between them.
For $x=4$, we have $\frac{43 x}{25}=6 \frac{12}{25}$ and $\frac{17 x}{9}=7 \frac{5}{9}$. Hence, in this case $\alpha=7$, and $\beta=4 \alpha+x=28+4=32$.
This is also the least possible value, because, if $x \geq 5$, then $\alpha>\frac{43 x}{25} \geq \frac{43}{5}>8$, and so $\beta>37$.
Hence the minimum possible value of $\beta$ is 32.