Four, (20 points) Find all positive numbers $t=\frac{b^{2}}{a^{2}}$, such that on the right branch of the hyperbola $\frac{\dot{x}^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a, b>0)$, there always exists a focal chord $l$ that subtends a right angle at the origin.
Let the right focus of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{t a^{2}}=1$ be $F(c, 0)$, where $c=a \sqrt{1+t}$. Let the focal chord $l$ intersect the right branch of the hyperbola at points $M\left(x_{1}, y_{1}\right)$ and $N\left(x_{2}, y_{2}\right)$.
From $O M \cdot O N=0$, we get
$x_{1} x_{2}+y_{1} y_{2}=0$.
If $t \leqslant 1$, then the inclination angle $\theta \leqslant$ $45^{\circ}$ of the asymptote $y=\sqrt{t} x$. The right branch of the hyperbola is contained within the angle region formed by the two asymptotes $y= \pm \sqrt{t} x$, with the vertex angle $\alpha \leqslant$ $90^{\circ}$. In this case, the angle subtended by any chord in the right branch of the hyperbola at the origin is less than $90^{\circ}$, which contradicts the given condition. Therefore, $t>1$.
(1) When the focal chord $l$ is perpendicular to the $x$-axis, if $\triangle O M N$ forms a right triangle with $M N$ as the hypotenuse, it is an isosceles right triangle. Thus, $M F \perp O F, M F=$ $O F$ (as shown in Figure 6), which means the point $M\left(x_{1}, y_{1}\right)$ satisfies $x_{1}=y_{1}$.
$$
\begin{array}{r}
\text { and } x_{1}=O F \\
=c=a \sqrt{1+t} . \\
\text { From } \frac{(1+t) a^{2}}{a^{2}}-
\end{array}
$$
$\frac{y_{1}^{2}}{t a^{2}}=1$, we get
$$
y_{1}=t a \text {. }
$$

Thus, $t=\sqrt{1+t}$.
Since $t>1$, we have
$$
t=\frac{1+\sqrt{5}}{2} \text {. }
$$
(2) When the focal chord $l$ is not perpendicular to the $x$-axis, the focal chord $l$ intersects the right branch of the hyperbola at two points with different $x$-coordinates. Let the equation of $l$ be $y=k(x-c)(k \neq 0)$.
Rewrite the equation of the hyperbola as $t x^{2}=y^{2}+t a^{2}$.
Then $t x^{2}=k^{2}(x-c)^{2}+t a^{2}$, which simplifies to
$$
\left(k^{2}-t\right) x^{2}-2 c k^{2} x+\left(k^{2} c^{2}+t a^{2}\right)=0
$$

Since equation (2) has two distinct positive roots $x_{1}$ and $x_{2}$, we have $k^{2}-t>0$.
From equation (2), we get
$$
x_{1}+x_{2}=\frac{2 c k^{2}}{k^{2}-t}, x_{1} x_{2}=\frac{k^{2} c^{2}+t a^{2}}{k^{2}-t} \text {. }
$$

According to equation (1), we have
$$
\begin{array}{l}
0=x_{1} x_{2}+y_{1} y_{2} \\
=x_{1} x_{2}+k^{2}\left(x_{1}-c\right)\left(x_{2}-c\right) \\
=\left(k^{2}+1\right) x_{1} x_{2}-k^{2} c\left(x_{1}+x_{2}\right)+k^{2} c^{2},
\end{array}
$$

which simplifies to $\left(k^{2}+1\right)\left(k^{2} c^{2}+t a^{2}\right)-2 k^{4} c^{2}+k^{2} c^{2}\left(k^{2}-t\right)$ $=0$.
Noting that $c^{2}=(1+t) a^{2}$, we get $k^{2}\left(t^{2}-t-1\right)=t$.
Since $t>1$ and $k^{2}>0$, we have $t^{2}-t-1>0$.
Therefore, $t>\frac{1+\sqrt{5}}{2}$.
Since $k^{2}>t$, from equation (3) we get $t^{2}-t-1<1$, i.e., $t^{2}-t-2<0 \Rightarrow t<2$.
Combining (1) and (2), the range of $t$ is
$$
\frac{1+\sqrt{5}}{2} \leqslant t<2 \text {. }
$$