11. (20 points) In the Cartesian coordinate system $x O y$, the parabola $C: x^{2}=y$. Suppose there exists a line $l$ passing through the point $P(0, t)$ that intersects the parabola $C$ at points $A$ and $B$, and there always exists a point $M$ on the parabola $C$ such that the line $l$ intersects the circumcircle of $\triangle M O P$, with the other intersection point being $Q$, and satisfies $\angle P M Q = \angle A M B = 90^{\circ}$. Find the range of positive values for $t$.
11. According to the problem, let $M\left(x_{0}, x_{0}^{2}\right)\left(x_{0} \neq 0\right)$, and the line $l: y=k x+t$.
By combining the equation of line $l$ with the parabola equation, we get $x^{2}-k x-t=0$.
Let $A\left(x_{1}, x_{1}^{2}\right), B\left(x_{2}, x_{2}^{2}\right)$. Then
$$
x_{1}+x_{2}=k, x_{1} x_{2}=-t \text {. }
$$

Given $\angle A M B=90^{\circ}$
$$
\begin{aligned}
\Rightarrow & \overrightarrow{M A} \cdot \overrightarrow{M B} \\
& =\left(x_{1}-x_{0}\right)\left(x_{2}-x_{0}\right)+\left(x_{1}^{2}-x_{0}^{2}\right)\left(x_{2}^{2}-x_{0}^{2}\right)=0 \\
\Rightarrow & 1+\left(x_{1}+x_{0}\right)\left(x_{2}+x_{0}\right)=0 \\
\Rightarrow & x_{0}^{2}+\left(x_{1}+x_{2}\right) x_{0}+x_{1} x_{2}+1=0 \\
\Rightarrow & x_{0}^{2}+k x_{0}-t+1=0 \\
\Rightarrow & k=-\frac{x_{0}^{2}-t+1}{x_{0}} .
\end{aligned}
$$

Let the circumcenter of $\triangle M O P$ be $T$. Thus, $T\left(x_{T}, \frac{t}{2}\right)$.
Since $\angle P M Q=90^{\circ}$, point $T$ lies on line $l$.
Also, point $T$ lies on the perpendicular bisector of $O M$, hence
$$
\begin{array}{l}
y_{T}-\frac{x_{0}^{2}}{2}=-\frac{1}{x_{0}}\left(x_{T}-\frac{x_{0}}{2}\right) \\
\Rightarrow x_{T}=\frac{x_{0}}{2}\left(x_{0}^{2}-t+1\right) .
\end{array}
$$

Then $k=\frac{y_{T}-y_{P}}{x_{T}-x_{P}}=\frac{-t}{x_{0}\left(x_{0}^{2}-t+1\right)}$.
From equations (1) and (2), we get
$$
\left(x_{0}^{2}-t+1\right)^{2}=t \text {. }
$$

Let $s=x_{0}^{2}+1$. Then the problem reduces to finding the real root greater than 1 of the equation
$$
f(s)=s^{2}-2 t s+t^{2}-t=0
$$

Thus,
$$
f(1)1 .\end{array}\right.$$
Solving this, we get $t^{2}-3 t+11$, which means
$$
t>\frac{3-\sqrt{5}}{2} \text {. }
$$

Therefore, $t \in\left(\frac{3-\sqrt{5}}{2},+\infty\right)$.