jmc

The three given points are labeled $A$, $B$, and $C$. The three possible values of the fourth point in the parallelogram are labelled $D_1$, $D_2$, and $D_3$, with $D_1$ being the opposite point of $A$, $D_2$ the opposite point of $B$, and $D_3$ the opposite point of $C$. The parallelogram $A{D}_{3}BC$ has the same perimeter as the parallelogram $ABC{D}_2$ by symmetry, so we disregard point $D_3$.

We will find the perimeter of $ABC{D}_2$. To calculate where point $D_2$ is, we notice that $A{D}_2$ must be parallel to the vertical segment $BC$, so the $x$ value of point $D_2$ must be $-1$. In addition, the length $A{D}_2$ must be equal to the length $BC$, which is 8. Thus, the $y$ value of point $D_2$ must be $-8$. So point $D_2$ is at $(-1,-8)$. The vertical segments of parallelogram $ABC{D}_2$ have length 8. To find the length of the diagonal segments $AB$ and $C{D}_2$, we use the distance formula between points $A$ and $B$: $AB=\sqrt{(-1-2{)}^2+(0-4{)}^2}=5$. Thus, the perimeter of this parallelogram is $8+8+5+5=26$.

We will find the perimeter of $AB{D}_{1}C$. To calculate where point $D_1$ is, we note that since figure $ABC$ is symmetric about the $x$-axis, $D_1$ must lie on the $x$-axis, so its $y$ value is 0. We also know that the diagonals in a parallelogram bisect each other, so in order for diagonal $A{D}_1$ to bisect $BC$ (which crosses the $x$-axis at $x=2$), the $x$ value of $D_1$ must be 5. So point $D_1$ is at $(5,0)$. In finding the perimeter, we note that all the sides are equal in length. Since we already found side $AB$ to have length 5, the entire perimeter is $5\cdot 4=20$.

Thus, the positive difference between the greatest perimeter and the smallest is $26-20=\boxed{6}$ units.