We have a stick of length $2n{}$ and a machine which cuts sticks of length $k\in\mathbb{N}$ with $k>1$ into two sticks with arbitrary positive integer lengths. What is the smallest number of cuts after which we can always find some sticks whose lengths sum up to $n{}$?
We show that at least $n$ cuts are required, and sufficient. First to see that at least these many cuts are required, consider cutting the stick $n-1$ times to obtain the lengths: \[n+1, 1, \dots, 1\]where there are $n-1$ ones. Clearly no subset sum of these sticks sums to a length of $n$, therefore we require at least $n$ cuts.
We now show that $n$ cuts are sufficient. Suppose the lengths of the new sticks are $a_1, a_2, \dots, a_{n+1}$. From a famous php, we know that there is a nonempty subset $S$ of sticks among $a_1, a_2, \dots, a_n$ with sum divisible by $n$. Since \[0 < \sum_{i \in S} a_i \leq 2n-a_{n+1} < 2n\]this forces said sum to be exactly $n$, whence $S$ does the job for us. $\square$