Write $S_0=0$ and $S_j=a_1+\cdots+a_j$. Rotate the sequence to start immediately after an index where $S_j$ is minimum. This only permutes the possible starting indices, so it preserves the number of successful shifts. The prefix sums of the rotated sequence are all nonnegative: before wrapping they are differences from the minimum, and after wrapping they have an additional $k$.

We may therefore assume that $S_j\ge0$ for all $j$, with $S_0=0$ and $S_n=k$.

For each integer $h\in\{0,\ldots,k-1\}$, let $t_h$ be the last index with $S_{t_h}=h$. Each such height occurs because the prefix sums go from $0$ to $k$ and every upward step is at most $1$. Since $S_n=k>h$, we have $t_h<n$.

Every later prefix sum is greater than $h$. A later value equal to $h$ contradicts the choice of $t_h$. A later value below $h$ would force another visit to $h$ before reaching $k$, because upward steps cannot skip an integer height. Thus the shift starting after $t_h$ has positive prefix sums before wrapping. After wrapping, its prefix sums have the form

$k+S_j-h\ge k-h>0.$

So these $k$ starting indices all work.

Conversely, suppose the shift starting after $t$ works. Its prefix ending at the original last element has sum $k-S_t>0$, so $0\le S_t<k$. No later prefix sum can equal $S_t$, since that would give a nonempty prefix of the shift with sum zero. Therefore $t$ is the last occurrence of its height and equals $t_{S_t}$.

There are exactly $k$ successful starting indices.

Source: Cycle Lemma, AtCoder Algorithm Lectures.