Solution:
Let the smallest value be $s$ and suppose it occurs $m$ times (with $m < n$). Then the values in the next stage are all at least $s$, and at most $m - 1$ equal $s$. So after at most $m$ iterations the smallest value is increased.

We can never reach a stage where all the values are equal, because if $(a_1 + a_2) / 2 = (a_2 + a_3) / 2 = \ldots = (a_{n - 1} + a_n) / 2 = (a_n + a_1) / 2$, then $a_1 + a_2 = a_2 + a_3$ and hence $a_1 = a_3$. Similarly, $a_3 = a_5$, and so $a_1 = a_3 = a_5 = \ldots = a_n$ ($n$ odd). Similarly, $a_2 = a_4 = \ldots = a_{n - 1}$. But we also have $a_n + a_1 = a_1 + a_2$ and so $a_2 = a_n$, so that all $a_i$ are equal. In other words, if all the values are equal at a particular stage, then they must have been equal at the previous stage, and hence at every stage.

Thus if the values do not start out all equal, then the smallest value increases indefinitely. But that is impossible, because the sum of the values is the same at each stage, and hence the smallest value can never exceed $(a_1 + \ldots + a_n) / n$.

Note that for $n$ even the argument breaks down because a set of unequal numbers can iterate into a set of equal numbers. For example: $1, 3, 1, 3, \ldots, 1, 3$.