10. (20 points) Through the focus $F_{0}$ 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 common chord of the circles $\odot M$ and $\odot N$ with diameters $AB$ and $CD$ is denoted as $l$. Find the minimum distance from point $M$ to the line $l$.

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 $EF$.
From the problem, we know $F_{0}(0,1)$.
Let $l_{1}: y=k_{1} x+1$.
10. As shown in Figure 4, draw perpendiculars from points \( M \) and \( N \) to the directrix \( y = -1 \), with the feet of the perpendiculars being \( E \) and \( F \) respectively.
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, \( x_{M} = \frac{x_{1} + x_{2}}{2} = 2 k_{1} \),
\( y_{M} = k_{1} x_{M} + 2 = 2 k_{1}^{2} + 2 \).
Therefore, \( M\left(2 k_{1}, 2 k_{1}^{2} + 2\right) \).
Similarly, \( N\left(2 k_{2}, 2 k_{2}^{2} + 2\right) \).
Notice that, \( y_{G} = -1 \),
\[
\begin{aligned}
x_{G} & = \frac{x_{E} + x_{F}}{2} = \frac{x_{M} + x_{N}}{2} = \frac{2 k_{1} + 2 k_{2}}{2} \\
& = k_{1} + k_{2} = 2 .
\end{aligned}
\]

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

Therefore, the slope of the common chord is \( -\frac{1}{k_{M N}} = -\frac{1}{2} \),
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
\[
\begin{aligned}
d & = \frac{\left|2 k_{1} + 2\left(2 k_{1}^{2} + 2\right)\right|}{\sqrt{5}} \\
& = \frac{4\left[\left(k_{1} + \frac{1}{4}\right)^{2} + \frac{15}{16}\right]}{\sqrt{5}} \geqslant \frac{3 \sqrt{5}}{4} .
\end{aligned}
\]

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