I also stumbled on the problem about how to exactly divide the bins, and could not find anything on the internet. So here is how I would formalize the problem to unambiguously assign the items to the $N$ desired bins.
Provided some multiset $L \in \mathbb{R}^n$, $n \in \mathbb{N}^+$ and the number of desired bins $N \in \mathbb{N}^+$.
Calculate the width $w \in \mathbb{R}$ like so (note that $w$ does not have to be a natural number):
\begin{equation*}
w := \frac{1}{N} \cdot \bigl(\max(L)-\min(L)\bigr)
\end{equation*}
Then calculate the intervals $I_1, \dots, I_N$:
\begin{equation*}
I_i :=
\begin{cases}
\Bigl[\min(L) + (i - 1) \cdot w,\ \min(L) + i \cdot w\Bigr),\ &i \in \{1, \dots, N-1\}\\
\Bigl[\min(L) + (N - 1) \cdot w,\ \min(L) + N \cdot w\Bigr], &i = N
\end{cases}
\end{equation*}
Alternatively, you can use:
\begin{equation*}
I_i :=
\begin{cases}
\Bigl[\min(L),\ \min(L) + w\Bigr], &i = 1\\
\Bigl(\min(L) + (i - 1) \cdot w,\ \min(L) + i \cdot w\Bigr],\ &i \in \{2, \dots, N\}
\end{cases}
\end{equation*}
Then fill each bin $B_1, \dots, B_N$ like so:
\begin{equation*}
B_i := \bigl\{x \in L \mid x \in I_i\bigr\} = L \cap I_i,\ i \in \bigl\{1, \dots, N\bigr\}
\end{equation*}
Pseudocode for inputs L[0, ..., n-1]
and N
desired bins, choosing the last interval as largest:
L := sorted(L)
width := (1/N) * (L[n-1] - L[0])
B[i] := [] for i in {0, ..., N-1}
i := 0
for x in L:
if x >= L[0] + i * width and i != N-1:
i := i + 1
B[i] := B[i] + [x]
With your example $L := \{5, 10, 11, 13, 15, 35, 50 ,55, 72, 92, 204, 215\}$ and $N := 3$, we get the following:
\begin{equation*}
w = \frac{1}{N} \cdot \bigl(\max(L) - \min(L)\bigr) = \frac{1}{3} \cdot (215 - 5) = 70\\
\begin{aligned}
I_1 &= \left[5, 75\right) \longrightarrow B_1 = \{5,10,11,13,15,35,50,55,72\}\\
I_2 &= \left[75, 145\right) \longrightarrow B_2 = \{92,204\}\\
I_3 &= \left[145, 215\right] \longrightarrow B_3 = \{215\}
\end{aligned}
\end{equation*}