Step 1: Write the curve in standard form.
The curve is $3y = 6x - 5x^3$, i.e., $y = 2x - \frac{5}{3}x^3$. Let $P = (a,\, 2a - \frac{5}{3}a^3)$ be a point on this curve.
Step 2: Find the slope of the tangent at $P$.
$\frac{dy}{dx} = 2 - 5x^2$. At $x = a$: slope of tangent $= 2 - 5a^2$.
Step 3: Find the slope of the normal at $P$.
Slope of normal $= -\frac{1}{2 - 5a^2}$, provided $2 - 5a^2 \ne 0$.
Step 4: Use the condition that the normal passes through $(0,0)$.
The slope of the line from $P = (a,\, 2a - \frac{5}{3}a^3)$ to $(0,0)$ is $\frac{2a - \frac{5}{3}a^3}{a} = 2 - \frac{5}{3}a^2$. Setting this equal to the normal slope: \[2 - \frac{5}{3}a^2 = -\frac{1}{2 - 5a^2}.\]
Step 5: Solve the equation.
Let $u = a^2$. Then $\left(2 - \frac{5}{3}u\right)(2 - 5u) = -1$. Expanding: $4 - 10u - \frac{10}{3}u + \frac{25}{3}u^2 = -1$. Multiply by 3: $12 - 30u - 10u + 25u^2 = -3$, so $25u^2 - 40u + 15 = 0$, i.e., $5u^2 - 8u + 3 = 0$, giving $(5u-3)(u-1) = 0$, so $u = 3/5$ or $u = 1$.
Step 6: Identify the positive integral value.
$a^2 = 1$ gives $a = \pm 1$; $a^2 = 3/5$ gives non-integers. The positive integral value is $a = 1$. \[\boxed{1}\]