To find the smallest positive integral value of \(a\) for which all the roots of the equation \(x^4 - ax^2 + 9 = 0\) are real and distinct, let's follow these steps:

First, observe that the given polynomial is a quadratic in terms of \(y = x^2\), so we can rewrite the equation as:

\(y^2 - ay + 9 = 0\)

For the roots of this quadratic equation in \(y\) to be real, the discriminant must be non-negative. The discriminant \(D\) of the equation \(y^2 - ay + 9 = 0\) is given by:

\(D = a^2 - 4 \times 1 \times 9\)

That simplifies to:

\(D = a^2 - 36\)

For the roots in \(y\) to be real and distinct, \(D\) must be positive:

\(a^2 - 36 > 0\)

This implies:

\(a^2 > 36\)

Taking the square root of both sides, we get:

\(a > 6\)

Since \(a\) must be a positive integer, the smallest integral value satisfying \(a > 6\) is \(a = 7\).

However, we need to ensure that with \(a = 7\), the condition of having all distinct, real roots of the original equation is met. The quadratic equation \(y^2 - 7y + 9 = 0\) will give two values of \(y\). For each \(y\), \(x^2 = y\) must give distinct \(x\), i.e., each \(y\) should be positive.

For \(a = 9\), the discriminant is zero:

\(9^2 - 36 = 45\)

This results in distinct real roots of the quadratic equation.

Therefore, the smallest positive integral value of \(a\) for which all the roots of the original polynomial are real and distinct is \(9\).