Three, (20 points) As shown in the figure,
the projection of the apex of the pyramid $S-ABCD$ onto the base is exactly the intersection point $O$ of the diagonals of the base. It is known that the height of the pyramid $S-ABCD$ is always $3$, $V_{s-ADO}=m^{2}$, $V_{s-BCO}=n^{2}\left(m, n \in R^{+}, m \neq n\right)$. When the volume of the pyramid $S-ABCD$ is minimized, what kind of quadrilateral is the base $ABCD$?
$$
\text { Three, } V_{S-A D O}=\frac{1}{3} S O \cdot S_{\triangle A D O}=S_{\triangle A D O}=m^{2} \text {. }
$$

Similarly, $S_{\triangle B C O}=n^{2}$.
Let $S_{\triangle D O C}=x, S_{\triangle A O B}=y$, then
$$
V_{S-A B C D}=\frac{1}{3} S O \cdot S_{\triangle B C D}=m^{2}+n^{2}+x+y \text {. }
$$

And $\frac{S_{\triangle A D O}}{S_{\triangle D O C}}=\frac{A O}{C O}=\frac{m^{2}}{x}=\frac{S_{\triangle A O B}}{S_{\triangle B O C}}=\frac{y}{n^{2}}$,
then $m^{2} n^{2}=x y$.
$$
\therefore \quad x+y \geqslant 2 \sqrt{x y}=2 m n \text {. }
$$

The equality holds only when $x=y=m n$, at this time $V_{s-A B C D}$ reaches its minimum value $(m+n)^{2}$. At this moment,
$$
\frac{A O}{O C}=\frac{S_{\triangle A O D}}{S_{\triangle D O C}}=\frac{m^{2}}{x}=\frac{m^{2}}{m n}=\frac{m}{n} .
$$

Also, $\frac{D O}{O B}=\frac{S_{\triangle D O C}}{S_{\triangle B D C}}=\frac{x}{n^{2}}=\frac{m n}{n^{2}}=\frac{m}{n}$,
$$
\therefore \frac{A O}{O C}=\frac{D O}{O B} \text {. }
$$

Thus, $A D / / B C$.
If $D C / / A B$, then $A B C D$ is a parallelogram, and $m^{2}$ $=n^{2}$, which leads to $m=n$, contradicting the given condition, so $D C / / A B$.
Therefore, quadrilateral $A B C D$ is a trapezoid.