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Two vertices of the first parallelogram are at $(0,c)$ and $(0,d)$.



The $x$-coordinates of the other two vertices satisfy $ax+c=bx+d$ and $ax+d=bx+c$, so the $x$-coordinates are $\pm (c-d)/(b-a)$. Thus the parallelogram is composed of two triangles, each of which has area $$9=\frac{1}{2}\cdot |c-d|\cdot \left|\frac{c-d}{b-a}\right|.$$It follows that $(c-d{)}^2=18|b-a|$.

By a similar argument using the second parallelogram, $(c+d{)}^2=72|b-a|$. Subtracting the first equation from the second yields $4cd=54|b-a|$, so $2cd=27|b-a|$. Thus $|b-a|$ is even, and $a+b$ is minimized when $\{a,b\}=\{1,3\}$. Also, $cd$ is a multiple of 27, and $c+d$ is minimized when $\{c,d\}=\{3,9\}$. Hence the smallest possible value of $a+b+c+d$ is $1+3+3+9=\boxed{16}$. Note that the required conditions are satisfied when $(a,b,c,d)=(1,3,3,9)$.