Is not hard!

$(\alpha - 3\beta)^{2} + 203(\alpha - 3)(\beta - 1) - 191\alpha\beta = 9$

$\alpha^{2} - 6\alpha\beta + 9\beta^{2} + 203\alpha\beta - 203\alpha - 609\beta + 609 - 191\alpha\beta = 9$

$\alpha^{2} + 6\alpha\beta + 9\beta^{2} - 203\alpha - 609\beta + 600 = 0$

$(\alpha + 3\beta)^{2} - 203(\alpha + 3\beta) + 600 = 0$

Let's say that $\alpha + 3\beta = x$

$x^2 - 203x + 600 = 0$

$(x - 200)(x - 3) = 0$

$x = 200$ or $x = 3$

Because $\alpha$ and $\beta$ are positive integers, $x = 3$ is impossible. We use $x = 200$ .

$\alpha + 3\beta = 200$

If $\beta = 1, \alpha = 197$

If $\beta = 2, \alpha = 194$

and forth, until $\beta = 66, \alpha = 2$

Count all possible values of $\alpha$

$2 + 5 + ... + 194 + 197 = 33(2 + 197) = \boxed{6567}$