Four. (17 points) As shown in Figure 4, given the line
$$
l: y=k x+2-4 k \text{ ( } k \text{ is a real number). }
$$
- (1) Prove that regardless of the value of $k$, the line $l$ always passes through a fixed point $M$, and find the coordinates of point $M$;
(2) If the line $l$ intersects the positive $x$-axis and $y$-axis at points $A$ and $B$ respectively, find the minimum value of the area of $\triangle A O B$.
(1) Let $k=1$, we get $y=x-2$; let $k=2$, we get $y=2x-6$. Solving these simultaneously, we get $x=4, y=2$. Therefore, the fixed point is $M(4,2)$.

Substituting the coordinates of point $M(4,2)$ into the equation of line $l$, we get $2=2$, which is an identity independent of $k$. Hence, regardless of the value of $k$, line $l$ always passes through the fixed point $M(4,2)$.
(2) Taking $x=0$, we get $OB=2-4k$ ($k<0$).
Taking $y=0$, we get $OA=\frac{4k-2}{k}$ ($k<0$).
Thus, the area of $\triangle AOB$ is
$$
S=\frac{1}{2} OA \cdot OB=\frac{1}{2} \cdot \frac{4k-2}{k}(2-4k) \quad (k<0).
$$

Converting the above expression into a quadratic equation in $k$, we get
$$
8k^2 + (S-8)k + 2 = 0.
$$

Since $k$ is a real number, we have $\Delta = (S-8)^2 - 64 \geq 0$, which simplifies to $S^2 - 16S \geq 0$, hence $S \geq 16$.
Substituting $S=16$ into equation (1), we get $k=-\frac{1}{2}$.
Therefore, when $k=-\frac{1}{2}$, $S_{\text{min}}=16$.