10. (20 points) Through the focus $F$ of the parabola $\Gamma: x^{2}=4 y$, draw two different lines $l_{1}$ and $l_{2}$ with slopes $k_{1}$ and $k_{2}$ respectively, and $k_{1}+k_{2}=2$. Line $l_{1}$ intersects the parabola $\Gamma$ at points $A$ and $B$, and line $l_{2}$ intersects the parabola $\Gamma$ at points $C$ and $D$. The circles $\odot M$ and $\odot N$ have diameters $AB$ and $CD$ respectively. The line containing the common chord of these circles is denoted as $l$. Find the minimum distance from point $M$ to the line $l$.
10. Draw perpendiculars from points $M$ and $N$ to the directrix $y=-1$, with feet of the perpendiculars being $E$ and $F$ respectively.

By the properties of the parabola, the directrix $y=-1$ is the common tangent of $\odot M$ and $\odot N$.

Let the common chord of the two circles intersect the common tangent at point $G$. Then, by the secant-tangent theorem, $G$ is the midpoint of $E F$.
By the problem, we know $F(0,1)$.
Let $l_{1}: y=k_{1} x+1$.
From $\left\{\begin{array}{l}x^{2}=4 y, \\ y=k_{1} x+1\end{array} \Rightarrow x^{2}-4 k_{1} x-4=0\right.$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then $x_{1}+x_{2}=4 k_{1}$.
Thus, $x_{M}=\frac{x_{1}+x_{2}}{2}=2 k_{1}$,
$y_{M}=k_{1} x_{M}+2=2 k_{1}^{2}+2$.
Hence, $M\left(2 k_{1}, 2 k_{1}^{2}+2\right)$.
Similarly, $N\left(2 k_{2}, 2 k_{2}^{2}+2\right)$.
Notice that, $y_{G}=-1$,
$$
x_{G}=\frac{x_{E}+x_{F}}{2}=\frac{x_{M}+x_{N}}{2}=k_{1}+k_{2}=2 .
$$

Thus, $G(2,-1)$.
Since the common chord is perpendicular to the line connecting the centers of the two circles,
$$
\begin{array}{l}
k_{M N}=\frac{\left(2 k_{2}^{2}+2\right)-\left(2 k_{1}^{2}+2\right)}{2 k_{2}-2 k_{1}} \\
=k_{1}+k_{2}=2,
\end{array}
$$

The line $l: y-(-1)=-\frac{1}{2}(x-2)$
$$
\Rightarrow x+2 y=0 \text {. }
$$

Thus, the distance from point $M$ to the line $l$ is
$$
\begin{aligned}
d & =\frac{\left|2 k_{1}+2\left(k_{1}^{2}+2\right)\right|}{\sqrt{5}} \\
& =\frac{4\left(\left(k_{1}+\frac{1}{4}\right)^{2}+\frac{15}{16}\right)}{\sqrt{5}} \\
& \geqslant \frac{3 \sqrt{5}}{4} .
\end{aligned}
$$

Therefore, when $k_{1}=-\frac{1}{4}$, the minimum distance from point $M$ to the line $l$ is $\frac{3 \sqrt{5}}{4}$.