, 2016

The answer is no. Regardless of the initial distribution all boxes will contain the same number of markers after finitely many steps. Moreover this is true for any number of boxes.

Denote by $x_n$ the number of markers in a certain box before step $n$, $n = 1, 2, \dots$. Suppose that $x_n = n$ for some $n$. Then by the rule of adding markers we have $x_{n+1} = n+1$, $x_{n+2} = n+2$ etc.; in other words the number of markers in that box equals the number of the oncoming step $l$ for each $l \ge n$. So, in order to prove that eventually all boxes contain the same number of markers, it is enough to show that for each box there exist a step $n$ such that $x_n = n$.

We use the following observation. Let a box C satisfy $x_i > l$ for some $l$, that is, the difference $d_i = x_i - l$ is positive. Then there is an $m \ge l$ such that C receives no marker at step $m$. Otherwise

$x_i$ increases by 1 at every step $m \ge l$, which means that $x_i + s$ is divisible by $l+s$ for all $s \ge 0$.

However this is impossible as $1 < \frac{x_i + s}{l+s} < 2$ for $s$ sufficiently large; it is enough to take $s > x_i - 2l$.

Let $m \ge l$ be the first step that adds no marker to C. Then the observation implies that the difference $d_{m+1} = x_{m+1} - (m+1) = x_m - (m+1)$ satisfies $d_{m+1} = d_l - 1$. If $d_{m+1} > 0$ then by the same reason there is a step $k > m$ with $d_k = d_m - 1$. Repeated applications of the same argument show that after finitely many steps there will be a step $s$ such that $d_s = 0$, that is, $x_s = s$.

Initially, before step 1, one has $x_1 \ge 1$ for each box C. This is ensured by the condition that every box contains a marker. If $x_1 = 1$ then $x_n = n$ holds for C already with $n=1$. Otherwise $x_i > 1$, so by the above $x_n = n$ will result after finitely many steps. As explained in the beginning, when this happens for all boxes, the numbers of markers in them will be the same.