Thanks to Fedor Petrov for an advice how to simplify the estimates.

We present a simple algorithm which finds an *optimal* partition, and then find some estimates on the number of parts in it. Call a partition satisfying the requirements *$m$-universal*.

**Step 1.** Consider any partition of $L$. For every $k$, denote by $S_k$ the sum of all parts not exceeding $k$. We claim that a partition is $m$-universal if and only if $(*)$ $S_k\geq (m-1)k$ whenever $(m-1)k\leq L$.

The `only if' part is easy: if $m-1$ persons want to get $k$ units each, all of them should get a combination of pieces accounted for $S_k$.

For the `if' part, distribute the pieces to the people, starting from the largest ones. When we want to give to someone a piece of length $k$, the total length of non-distributed pieces is at least $S_{k-1}+k\geq (m-1)(k-1)+k=m(k-1)+1$, so there is a person who needs at least $k$; give them the piece and continue. The claim is proved.

**Step 2.** Now we present an algorithm which finds one of optimal $m$-universal partitions.

Define the sequence $a_1,a_2,\dots$ inductively as follows. On the $i$th step, when $a_1,\dots,a_i$ are already defined, find the minimal $k$ such that $\Sigma_i:=a_1+\dots+a_i<(m-1)k$ and set $a_{i+1}=k$ (thus, in partivular, $a_1=1$). In other words,
$$
a_{i+1}=\left\lfloor\frac{\Sigma_i}{m-1}\right\rfloor+1.
$$
For any $i$ we have
$$
(a_i-1)(m-1)\leq \Sigma_{i-1}<\Sigma_i<a_{i+1}(m-1),
$$
whence $a_i\leq a_{i+1}$. So the sequence is non-decreasing.

Now, cut from $L$ the pieces of lengths $a_1,a_2,\dots$ until the remainder does not exceed the next term of the sequence. This way, we obtain a partition $\mathcal P$ into pieces $a_1,\dots,a_{s-1},b$, where $0<b\leq a_s$. We claim that we obtain an optimal $m$-universal partition.

By the definition of $a_i$, the partition $\mathcal P$ satisfies $(*)$ (indeed, if the partition violates $(*)$ for some $k$, why did we set some term of $(a_i)$ to be larger than $k$?). Consider now any $m$-universal partition $\mathcal Q$ into parts $c_1\leq c_2\leq \dots$. We claim that $c_i\leq a_i$ for all $i\leq s-1$; this yields that $c_1+\dots+c_{s-1}<L$, so $\mathcal Q$ contains at least $s$ parts, and hence $\mathcal P$ is optimal.

Indeed, assume that $c_i>a_i$ for some (minimal) $i$. By the definition of $a_i$, we have $a_1+\dots+a_{i-1}<(m-1)a_i$ and hence $c_1+\dots+c_{i-1}<(m-1)a_i$ as well. But then in $\mathcal Q$ we have $S_{a_i}<(m-1)a_i$, so $\mathcal Q$ violates $(*)$.

Thus, our algorithm indeed provides an optimal partition.

**Step 3.** Notice that $\Sigma_n$ is the largest value of $L$ for which we can survive with $n$ pieces. Recall that $a_n=\lfloor \Sigma_{n-1}/(m-1)\rfloor+1$, so
$$
\frac {m\Sigma_{n-1}+1}{m-1}\leq \Sigma_n\leq\frac m{m-1}\Sigma_{n-1}+1.
$$
This rewrites as
$$
\frac m{m-1}(\Sigma_{n-1}+1)\leq \Sigma_n+1
\quad\text{and}\quad
\Sigma_n+(m-1)\leq \frac m{m-1}\left(\Sigma_{n-1}+(m-1)\right).
$$
Therefore,
$$
\left(\frac m{m-1}\right)^{n-t}(\Sigma_t+1)-1\leq \Sigma_n
\leq \left(\frac m{m-1}\right)^{n-t}(\Sigma_t+(m-1))-(m-1).
$$

Now, we may choose an appropriate value of $t$. Setting $t=m$ (where $\Sigma_m=m+1$), we get
$$
\left(\frac m{m-1}\right)^{n-m}(m+2)-1\leq \Sigma_n\leq
2m\left(\frac m{m-1}\right)^{n-m}-(m-1).
$$
This, in view of $\Sigma_{s-1}<L\leq \Sigma_s$, yields the estimates
$$
m+\log_{m/(m-1)}\frac{L+m-1}{2m}\leq s \leq m+1+\log_{m/(m-1)}\frac{L+1}{m+2}
$$
which differ only by $O(m)$ whenever $L\geq m$ (as expected).

**Remark 1.** One can see from Step 3 that there is another way of constructing an optimal partition. Namely, one may say that the largest block in the partition is $\lceil L/m\rceil$ (it cannot be larger!), set $L'=L-\lceil L/m\rceil$, and apply the procedure to $L'$. An argument similar to the `if' part of Step 1 shows that the obtained partition is $m$-universal. Using the above inequalities, one can show it is optimal.

This example readily yields $s\leq \log_{m/(m-1)}L+1$.

**Remark 2.** One may try to improve the bound by a smarter choice of $t$. E.g., with some more effort (estimating the length of the piece containing each of the first $(m-1)^2$ cubes) one can obtain
$$
\Sigma_{(m-1)\lfloor\ln m\rfloor}\leq (m-1)^2
\leq \Sigma_{m+m\lceil\ln(m-2)\rceil},
$$
which implies
$$
\left(\frac m{m-1}\right)^{n-m(2+\ln(m-2))}((m-1)^2+1)-1\leq \Sigma_n\leq
\left(\frac m{m-1}\right)^{n-(m-1)(\ln m-1)}m(m-1)-(m-1).
$$
This yields more complicated estimates
$$
\log_{m/(m-1)}\frac{L+(m-1)}{m(m-1)}+(m-1)\ln(m-1)\leq s<\log_{m/(m-1)}\frac{L+1}{m^2-2m+2}+1+m(2+\ln(m-2)),
$$
but the difference is still $O(m)$.

numberof partitions accomplishing this is given by the OEIS sequences A236970 which includes links to the analogous sequences for $m=4,5$. You want the number of parts in the minimal length partition that's count. There's a link to Haskell code that might be helpful. For $m=2$, you found the length of the smallest complete partition A126796 $\endgroup$