Cal people standing on a track $A_{1}, A_{2}...A_{n}$. Imagine regular $n$-gon on this track $B_{1}, B_{2}...B_{n}$. Consider $n-1$ moves ( $A_{1};A_{n}$ ) ; ( $A_{2};A_{n}$ ) ... ( $A_{n-1};A_{n}$ ) Such that $A_{i}$ will go the shortest arc between $A_{i}$ and $B_{i}$ and end up at $B_{i}$ for $i \leq n-1 $. Define $d_{i}$ as the diffrence between the lengths of $A_{i}$ went clockwise and anticlockwise. It is easy to see that $d_{1} + d_{2}+...+d_{n}=0$ (1) If $A_{n}$ ended at $B_{n}$ then we found desired $n-1$ moves . Else define the shortest length from the final position of $A_{n}$ to $B_{n}$ as $l$. ( $l$ can be negative ) and $ a = \frac{l}{n}$. Now we look at moves ( $A_{1};A_{n}$ ) ; ( $A_{2};A_{n}$ ) ... ( $A_{n-1};A_{n}$ ) such that $d'_{i}=d_{i} - a $ for $i \leq n-1$ And since (1) $d'_{n}= d_{n} + (n-1)a$ And now it is easy to see that the final positions of $A_{i}$ form $n$-gon $B'_{i}$ which is transformed from $B_{i}$ by moving each vertex on the distance $-a$

Suppose the clockwise or anticlockwise moving guy meets another person during his walk: can he move right through the other guy, or is the move blocked at this point?

No, he is not blocked.

Identify the circular track with the unit circle (centered at the origin $O$), and associate with every man $M$ on the track the angle $\alpha(M)$ that the vector $OM$ encloses with the positive part of the $x$-axis. For a situation with $n$ men $M_1,\ldots,M_n$ define the potential function $\sum_i \alpha(M_i) \pmod{2\pi}$. In the initial situation, there are two consecutive men $M_a$ and $M_b$ along the track such that the angle $\angle M_aOM_b$ is at least $2\pi/n$. First we embed the initial situation on the unit circle such that $M_a$ is standing at the intersection point $X$ of unit circle and positive $x$-axis. Then we slowly rotate the situation by an angle of $2\pi/n$. As during this rotation no man crosses point $X$, the rotation increases the angle of every man by $2\pi/n$ and thus increases the potential by $2\pi$. Hence somewhere during the rotation process, the potential must become zero; and that's the good embedding that we choose for the initial situation. For the final situation, we choose a regular $n$-gon $P_1,\ldots,P_n$ on the unit circle that has one of its vertices in point $X$. It is easy to see that the potential value of this final situation is zero. Now we perform the following $n-1$ moves: In the $k$th move ($k=1,\ldots,n-1$), the $k$th man moves from his current position clockwise to the vertex $P_k$ of the regular polygon. Simultaneously, the $n$th and last man moves the same distance counter-clockwise. As a move increases the angle of the counter-clockwise moving man by the same amount as it decreases the angle of the clockwise moving man, the value of the potential function remains constant throughout the game. After the $(n-1)$th move, the first $n-1$ men are standing in the points $P_1,P_2,\ldots,P_{n-1}$. Since the potential is still zero, the only possible position for the last man is the point $P_n$.