This may be a question with a well known answer, but I've been thinking on it for two days, and can't quite come up with a satisfactory answer.
Consider the problem of dividing $p n$ bins numbered $1$ through $pn$ into $p m + 1$ segments by placing $pm$ balls. If we let $k = pn - pm \bmod pm+1$, then we can show that we may attain a placement of the $pm$ balls such that there are exactly $(pm+1) - k$ segments of empty bins of length $\lfloor \frac{pn-pm}{pm+1} \rfloor$ and $k$ segments of empty bins of length $\lceil \frac{pn-pm}{pm + 1} \rceil$.
Here's the tricky question: can we accomplish this task while ensuring that there are exactly $m$ balls in each interval $[(j-1)n + 1, jn]$, for $j \in \{1, 2, \dots, p\}$?
Every concrete example I work through answers in the affirmative, but I can't seem to get an algorithmic way of doing it, or mathematical proof that one such arrangement exists.
For particular examples, it always seems to work. See an example below with $p = 2,$ $n = 7,$ $m = 2.$
1| |X| | |X|1|X| | |X| | |
where 1 denotes the beginning of a new 'period', and | denotes the 'wall' of a bin, and X denotes a ball. Note that 1 and | both denote 'walls' of bins.