Two thieves stole an open chain with $2k$ white beads and $2m$ black beads. They want to share the loot equally, by cutting the chain to pieces in such a way that each one gets $k$ white beads and $m$ black beads. What is the minimal number of cuts that is always sufficient?
1. **Labeling the Beads:**
   - We label the beads modulo $2(k+m)$ as $0, 1, 2, \ldots, 2(k+m)-1$.

2. **Considering Intervals:**
   - We consider intervals of length $k+m$ of the form $[x, x+(k+m)-1]$.

3. **Analyzing the Number of White Beads:**
   - As $x$ changes by one, the number of white beads in the interval changes by at most one.
   - Let $\ell$ be the number of white beads in the interval when $x=0$.
   - When $x=k+m$, the interval will have $2k-\ell$ white beads.

4. **Applying the Discrete Intermediate Value Theorem:**
   - By the discrete intermediate value theorem, since $\min(\ell, 2k-\ell) \le k \le \max(\ell, 2k-\ell)$, there exists some $x$ such that the interval $[x, x+(k+m)-1]$ has exactly $k$ white beads.
   - Since the interval has $k+m$ total beads, it must also have $m$ black beads.

5. **Complement Interval:**
   - The complement of this interval will also have $k$ white beads and $m$ black beads.

6. **Conclusion on Number of Cuts:**
   - Therefore, we can achieve the desired distribution with two cuts.
   - Clearly, one cut is insufficient because the chain will still be in one piece.

Thus, the minimal number of cuts that is always sufficient is two.

The final answer is $\boxed{2}$