Four. (20 points) Given the ellipse $\frac{x^{2}}{2}+y^{2}=1$ with its upper vertex $B$ and right focus $F$, a line $l$ intersects the ellipse at points $M$ and $N$. Determine whether there exists a line $l$ such that point $F$ is the
(1) centroid; (2) orthocenter of $\triangle B M N$.
If it exists, find the equation of the corresponding line $l$; if not, explain the reason.
(1) Assume there exists a line $l$ that satisfies the condition, making point $F$ the centroid of $\triangle B M N$.
Let $M(\sqrt{2} \cos \alpha, \sin \alpha), N(\sqrt{2} \cos \beta, \sin \beta)$.
According to the problem, we have
$$
\left\{\begin{array}{l}
\sqrt{2} \cos \alpha+\sqrt{2} \cos \beta+0=3 \times 1, \\
\sin \alpha+\sin \beta+1=3 \times 0 .
\end{array}\right.
$$

Simplifying, we get $\left\{\begin{array}{l}\cos \alpha+\cos \beta=\frac{3}{\sqrt{2}}, \\ \sin \alpha+\sin \beta=-1 .\end{array}\right.$
Squaring and adding the two equations, we get $2+2 \cos (\alpha-\beta)=\frac{11}{2}$, which simplifies to
$$
\cos (\alpha-\beta)=\frac{7}{4}>1,
$$

which is a contradiction. Therefore, point $F$ cannot be the centroid of $\triangle B M N$.
(2) Assume there exists a line $l$ that satisfies the condition, making point $F$ the orthocenter of $\triangle B M N$. Given $B(0,1), F(1,0)$, the slope of line $B F$ is -1. Thus, since $B F \perp M N$, the slope of line $l$ is:
Let the equation of line $l$ be $y=x+b$.
From $\left\{\begin{array}{l}\frac{x^{2}}{2}+y^{2}=1, \\ y=x+b\end{array}\right.$, eliminating $y$, we get
$$
3 x^{2}+4 b x+2\left(b^{2}-1\right)=0 \text {. }
$$

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. According to Vieta's formulas, we have
$$
x_{1}+x_{2}=-\frac{4 b}{3}, x_{1} x_{2}=\frac{2 b^{2}-2}{3} \text {. }
$$

Since $N F \perp B M$, we have $N F \cdot B M=0$.
And $N F=\left(1-x_{2},-y_{2}\right), B M=\left(x_{1}, y_{1}-1\right)$, so $\boldsymbol{N} \boldsymbol{F} \cdot \boldsymbol{B} \boldsymbol{M}=\left(1-x_{2}\right) x_{1}-y_{2}\left(y_{1}-1\right)$
$$
\begin{array}{l}
=x_{1}+y_{2}-x_{1} x_{2}-y_{1} y_{2} \\
=x_{1}+x_{2}+b-x_{1} x_{2}-\left(x_{1}+b\right)\left(x_{2}+b\right) \\
=-2 x_{1} x_{2}+(1-b)\left(x_{1}+x_{2}\right)+b-b^{2} \\
=-2 \cdot \frac{2 b^{2}-2}{3}+(1-b)\left(-\frac{4 b}{3}\right)+b-b^{2}=0 .
\end{array}
$$

Solving, we get $b=1$ or $b=-\frac{4}{3}$.
When $b=1$, point $B$ is the intersection point of line $l$ and the ellipse, which satisfies the condition. Therefore, the equation of the required line $l$ is $y=x-\frac{4}{3}$.