$$
\begin{array}{l}
M=\left\{(x, y) \left\lvert\, y \geqslant \frac{1}{4} x^{2}\right.\right\}, \\
N=\left\{(x, y) \left\lvert\, y \leqslant-\frac{1}{4} x^{2}+x+7\right.\right\}, \\
\left.D_{r}\left(x_{0}, y_{0}\right)=|(x, y)|\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2} \leqslant r^{2}\right\} .
\end{array}
$$

Find the largest $r$ such that $D_{r}\left(x_{0}, y_{0}\right) \subset M \cap N$.
2. Let $f(x)=\frac{1}{4} x^{2}$,
$g(x)=-\frac{1}{4} x^{2}+x+7=-\frac{1}{4}(x-2)^{2}+8$.
Then $M \cap N \neq \varnothing \Leftrightarrow f(x) \leqslant g(x)$ has real solutions $\Leftrightarrow x^{2}-2 x-14 \leqslant 0$ has real solutions.
Therefore, $M \cap N \neq \varnothing$.
In the Cartesian coordinate system, the vertices of $y=f(x)$ and $y=g(x)$ are $O(0,0)$ and $A(2,8)$, respectively, and the two parabolas are symmetric about the point $B(1,4)$.

Thus, the figure $T$ corresponding to $M \cap N$ in the coordinate system is symmetric about the point $B$, and the largest radius circle contained in $T$ should be the incircle of $T$.
We now prove: the center of this circle is $B$.
Otherwise, let the center of this circle be $B_{1}(\neq B)$, with radius $r$. Take the point $B_{1}$'s symmetric point about $B$ as $B_{2}$. Then the circle with $B_{2}$ as the center and $r$ as the radius is also an incircle of $T$.
Draw two external common tangents of the two circles.
Since $T$ is convex, the region enclosed by the above two external common tangents and the two circles is inside $T$.

Thus, the circle with $B$ as the center and $r$ as the radius is inside $T$ (excluding the boundary). Therefore, the radius of the incircle with $B$ as the center is greater than $r$, which is a contradiction.

Since $T$ is symmetric about the point $B$, the incircle $\odot B$ is the circle with $B$ as the center and tangent to $y=\frac{1}{4} x^{2}$. Let the radius of $\odot B$ be $r_{0}$.
Suppose $\odot B$ is tangent to $y=\frac{1}{4} x^{2}$ at point $C(a, b)$. Then
$b=\frac{a^{2}}{4}$.
The tangent line $l$ of $y=\frac{1}{4} x^{2}$ at point $C$ is
$y=\frac{a}{2}(x-a)+b$.
Clearly, $B C \perp l$, i.e.,
$\frac{b-4}{a-1} \cdot \frac{a}{2}=-1$.
Substituting equation (1) into equation (2) and simplifying, we get $a^{3}-8 a-8=0$.
Solving, we get $a=-2$ or $1 \pm \sqrt{5}$.
Also, $r_{0}=|B C|=\sqrt{\left(\frac{a}{2}\right)^{2}+1} \cdot|1-a|$, substituting the values of $a$ we find that the minimum value of $r_{0}$ is the radius of $\odot B$, i.e., when $a=1+\sqrt{5}$, $r_{0}=\sqrt{\frac{25-5 \sqrt{5}}{2}}$.
Therefore, $r_{\text {max }}=r_{0}=\sqrt{\frac{25-5 \sqrt{5}}{2}}$.