Let $$S$$ be the set of positive integral values of $$a$$ for which $$\frac{ax^2 + 2(a+1)x + 9a + 4}{x^2 - 8x + 32} < 0, \forall x \in \mathbb{R}$$. Then, the number of elements in $$S$$ is:

For $$x^2 - 8x + 32$$, the discriminant $$D = (-8)^2 - 4(32) = 64 - 128 = -64$$.

Since $$D < 0$$ and the leading coefficient is positive, the denominator is always positive for all $$x$$.

For the fraction to be $$< 0$$, the numerator must be always negative:

$$ax^2 + 2(a+1)x + 9a + 4 < 0$$

This requires:

1. $$a < 0$$ (parabola must open downward)

2. $$D < 0$$ (no real roots

The question asks for positive integral values of $$a$$. 

Since condition 1 requires $$a < 0$$, there are no positive integers that satisfy this.

Number of elements in $$S = \mathbf{0}$$ (Option B)