At the AoPS office, mice vary inversely with cats, that is, $\text{mice}=\frac{k}{\text{cats}}$, for some value of $k$. When there are $3r-19$ cats, there are $2r+1$ mice, and when there are $6r+2$ mice, there are $4r+3$ cats. Find $k$.
From the first statement, we know that \[2r+1=\frac{k}{3r-19}.\]From the second statement, we know that \[4r+3=\frac{k}{6r+2}.\]Since $\dfrac{k}{3r-19}=\dfrac{k}{6r+2}$, \begin{align*} 2r+1&=4r+3 \\ -2r&=2 \\ r&=-1.\end{align*}The values $r=-1$ satisfy both original statements, so \[2(-1)+1=4(-1)+3=-2+1=-1.\]At $r=-1$, we can calculate \[2r+1=2(-1)+1=-1.\]Thus, $k=(2(-1)+1)(3(-1)-19)=-1(-4-19)=23(-1)=\boxed{-23}$, which is our answer.
To find the value of $k$, we need to set up a system of equations using the given information.
First, we have the equation $\text{mice} = \frac{k}{\text{cats}}$.
Using the information that when there are $3r-19$ cats, there are $2r+1$ mice, we can write the equation as:
$2r+1 = \frac{k}{3r-19}$
Simplifying this equation, we have:
$(2r+1)(3r-19) = k$
Expanding the left side of the equation:
$6r^2 - 38r + 3r - 19 = k$
$6r^2 - 35r - 19 = k$
From the information that when there are $6r+2$ mice, there are $4r+3$ cats, we can set up another equation:
$4r+3 = \frac{k}{6r+2}$
Simplifying this equation, we have:
$(4r+3)(6r+2) = k$
Expanding the left side of the equation:
$24r^2 + 14r + 18r + 6 = k$
$24r^2 + 32r + 6 = k$
Now, we have two equations:
$6r^2 - 35r - 19 = k$
$24r^2 + 32r + 6 = k$
We can set these two equations equal to each other and solve for $r$:
$6r^2 - 35r - 19 = 24r^2 + 32r + 6$
$18r^2 - 67r - 25 = 0$
Now, we can solve this quadratic equation for $r$.