Four. (20 points) Given the ellipse $C: \frac{x^{2}}{4}+y^{2}=1$ and a fixed point $P(t, 0)(t>0)$, a line $l$ with a slope of $\frac{1}{2}$ passes through point $P$ and intersects the ellipse $C$ at two distinct points $A$ and $B$. For any point $M$ on the ellipse, there exists $\theta \in[0,2 \pi]$, such that $O M=\cos \theta \cdot O A+\sin \theta \cdot O B$ holds. Try to find the value of the real number $t$ that satisfies the condition.
Let $M(x, y)$ and points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. The equation of line $AB$ is $x-2 y-t=0$.
From the cubic equation, we get
$2 x^{2}-2 t x+t^{2}-4=0$ and $8 y^{2}+4 t y+t^{2}-4=0$.
Thus, $x_{1} x_{2}=\frac{t^{2}-4}{2}$ and $y_{1} y_{2}=\frac{t^{2}-4}{8}$.
Given $O M=\cos \theta \cdot O A+\sin \theta \cdot O B$, we have
$$
\begin{array}{l}
x=x_{1} \cos \theta+x_{2} \sin \theta, \\
y=y_{1} \cos \theta+y_{2} \sin \theta .
\end{array}
$$

Using $4=x^{2}+4 y^{2}$
$$
\begin{aligned}
= & \left(x_{1} \cos \theta+x_{2} \sin \theta\right)^{2}+4\left(y_{1} \cos \theta+y_{2} \sin \theta\right)^{2} \\
= & \left(x_{1}^{2}+4 y_{1}^{2}\right) \cos ^{2} \theta+\left(x_{2}^{2}+4 y_{2}^{2}\right) \sin ^{2} \theta+ \\
& 2 \sin \theta \cdot \cos \theta \cdot\left(x_{1} x_{2}+4 y_{1} y_{2}\right) \\
= & 4\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+2 \sin \theta \cdot \cos \theta . \\
& \left(x_{1} x_{2}+4 y_{1} y_{2}\right),
\end{aligned}
$$

We get $2 \sin \theta \cdot \cos \theta \cdot\left(x_{1} x_{2}+4 y_{1} y_{2}\right)=0$.
Since $\theta \in[0,2 \pi]$ is arbitrary, we know
$$
x_{1} x_{2}+4 y_{1} y_{2}=\frac{t^{2}-4}{2}+4 \times \frac{t^{2}-4}{8}=t^{2}-4=0 \text {. }
$$

Solving for $t$, we get $t=2$.
Substituting back, the condition is satisfied.
Therefore, the value of $t$ that satisfies the condition is 2.