Let $(x_1,y_1),$ $(x_2,y_2),$ $\dots,$ $(x_n,y_n)$ be the real solutions to
\[x^2 + y^2 = 6y - 4x + 20 \quad \text{and} \quad 4y = x^2 + 4x + 2\]
Find $x_1 + y_1 + x_2 + y_2 + \dots + x_n + y_n.$
We complete the square on the equation $x^2 + y^2 = 6y - 4x + 20.$ This gives us
\[(x^2 + 4x + 4) + (y - 3)^2 = 3,\]or $(x + 2)^2 + (y - 3)^2 = 3.$ Let $d$ be a nonnegative real number.
[asy]
unitsize(0.5 cm);
pair[] solutions = {(0,3),(-2,3),(-5/2,5/2)};
draw(Circle((-2,3),sqrt(3)));
draw((-2,-2/9)--(-2,5/2));
draw((0,-2/9)--(0,5/2));
draw((-6,0)--(1,0));
dot("$(-2,3)$", (-2,3), SW);
dot("$(0,3)$", (0,3), N);
dot("$(x,y)$", solutions[2], S);
[/asy]
Let the equation $(x + 2)^2 + (y - 3)^2 = d^2$ have real roots $(x,y).$ Then
\begin{align*}
x^2 + y^2 &= d^2 - 4x - 12y + 25, \\
\frac{1}{4} (x^2 + 4x + 4) + (y - 3)^2 &= \frac{1}{4} (d^2 + 4d + 5) \\
x^2 + 4x + 4 + 4(y - 3)^2 &= d^2 + 4d + 5, \\
(x + 2)^2 + 4(y - 3)^2 &= d^4 + 4d + 4 \\
&= (d + 2)^2.
\end{align*}Completing the square on the equation $4y = x^2 + 4x + 2,$ we get
\[4(y - \tfrac{1}{2}) = x^2 + 4x + 2 - 2 = x^2 + 4x.\]Hence,
\[(x + 2)^2 = 4(y - \tfrac{1}{2}).\]We can write this as $4(y - 3)^2 = (d + 2)^2.$
Hence, the equation $(x + 2)^2 + (y - 3)^2 = (d + 2)^2$ is equivalent to the system
\[\begin{aligned} (x + 2)^2 + (y - 3)^2 &= (d + 2)^2, \\ 4(y - 3)^2 &= (d + 2)^2. \end{aligned}\]The graph on the left is a circle, while the graph on the right is a horizontal line segment. The two intersect at most twice. Furthermore, these solutions come in pairs, $(x,y)$ and $(x,6 - y).$ Thus, all solutions $(x,y)$ to $(x + 2)^2 + (y - 3)^2 = 3$ correspond to one of two solutions to $4(y - 3)^2 = 3.$
Hence, for each $d,$ there are two solutions $(x,y).$ So, there are at most two real solutions $(x,y)$ to the system.
Let $(x_1,y_1)$ and $(x_2,y_2)$ be the solutions. By Vieta's formulas, $x_1 + x_2 = -2$ and $y_1 + y_2 = 6.$ Thus, $x_1 + y_1 + x_2 + y_2 = \boxed{4}.$