Josephine problem So the problem is
Suppose there are $2n$ people in a circle; the first $n$ are “good
guys” and the last $n$ are “bad guys.” Show that there is always an integer $m$ (depending on $n$) such that, if we go around the circle executing every $m$th person, all the bad guys are first to go.
I can see its true by simply visualizing $n=8, k=2$ so that if prisoners were at positions $2,4,6,8$ then all the bad guys would be dead. But how can I generalize and make proof work for everything?
 A: You are working on a particular variant of the Josephus Problem. Executing every $m^{\text{th}}$ person in a circle of $n$ defines the Josephus Permutation $$J_{n,m}=\begin{pmatrix} 1&2&3&\cdots &n\\ a_1&a_2&a_3&\cdots&a_n\end{pmatrix}$$ where $a_i$ is the $i^{\text{th}}$ person to be killed.
Now, there is a simple condition via the Chinese Remainder Theorem for an arbitrary permutation to be a Jospehus Permutation. Observe that a permutation $J_{n,m}$ can be completely described by an $n$-tuple of numbers ${\bf b}:=(b_1,b_2,\cdots,b_n)$ such that $b_i\leq n+1-i$ denotes the number of steps, clockwise, we have to make to kill $a_i$ after we just killed $a_{i-1}$. For instance $$J_{8,m}=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 4&1&6&2&5&7&3&8\end{pmatrix}\quad \text{ gives }\quad {\bf b}=(4,5,4,3,2,1,2,1)$$ since we need $4$ steps to the right to kill the $4^{\text{th}}$ person, then $5$ steps to the right (which completes the circle and starts back at the beginning) to kill the $1^{\text{st}}$, then only $4$ steps to kill the $6^{\text{th}}$ becuse we don't count the $4^{\text{th}}$ anymore, etc... Therefore, to be able to find a suitable $m$ that realizes this permutation as $J_{8,m}$, we need a solution of $$ \begin{alignat*}{1} m&=4\ \operatorname{mod}8\\ m&=5\ \operatorname{mod}7\\m&=4\ \operatorname{mod}6\\&\cdots \\m&=2\ \operatorname{mod}2\\ m&=1\ \operatorname{mod}1 \end{alignat*}$$
It is easy to check, using the Chinese Remainder Theorem, that such a solution does indeed exists. However, not all permutations are realized as Josephus Permutations:
Indeed, there can only be $\operatorname{lcm}(1,2,\cdots,n)$-many such systems  that have a solution (and hence equally many Josephus Permutations) but the Prime Number Theorem implies that $$\operatorname{lcm}(1,2,\cdots,n)\sim e^{n(1+o(1))}$$ which by Stirling's formula grows much slower than $n!$ ...
Now, for your particular question, you only need a Josephus permutation $J_{2n,m}$ such that the first $n$ people to die are a rearrangement of the numbers $n+1,n+2,\cdots,2n$. We can be even more precise:

Proposition The Permutation $$\sigma=\begin{pmatrix} 1&2&3&\cdots &n-1&n\\ n&n-1&n-2&\cdots& 2&1 \end{pmatrix}$$ is a Josephus permutation.
Indeed, just notice that the corresponding ${\bf b}$ is just $(n,n-1,n-2,\cdots,2,1)$ and then, of course, the system $b_i=0\ \operatorname{mod}(n+1-i)$ always has a solution.

The problem becomes more interesting once one tries to study other aspects of the permutations $J_{n,m}$. Traditionally, people attempt to find algorithms (some times there exist formulas! ) that give the last (one or two) survivors, ie $J_{n,m}(n)$, etc.
For references, consult MathSciNet with the term "Josephus" or have a look at these particular papers:


*

*The feline Josephus Problem.

*An application of Fourier transforms on finite abelian groups to an enumeration arising from the Josephus problem.

*The Josephus problem: once more around.

*The Josephus problem.
A: Are we allowed to start at any point in the circle? Anyway we can view it as using clock arithmetic or "modulus" or $\mathbb{Z}_{2n}$. Another approach is by matrix representation of groups. Then we can write the problem as $$ \left(P^l\sum_{k=1}^n P^{kt} \right)[1,0,\cdots,0]^T = v$$ where v is a binary vector being 1 in the bad guys positions and P being generator for the 1-step forward permutation matrix and $t,l$ some integers deciding where to start and how many steps for between each "shot".
