Abstract: In classical fixed-interval scheduling, each job of a given set must be processed during a prescribed time interval. The goal is to assign each job to exactly one machine such that no two jobs assigned to the same machine overlap in their interiors, and the number of machines used is minimized. Without further constraints this is interval graph coloring and is solvable in polynomial time through greedy approaches. We study a variant in which every used machine must remain idle during a contiguous \emph{break} of prescribed length $x$ somewhere in the scheduling horizon. We show that this additional constraint makes the problem hard, except when $x\le1$. First, for every fixed $x\ge2$, deciding whether $k$ machines suffice for the assignment of a given set of jobs is NP-complete, even when all coordinates are bounded linearly in the number of jobs, so machine minimization is strongly NP-hard. Second, for $x=1$, we give a polynomial-time algorithm that solves the problem exactly. Finally, we give a deterministic polynomial-time algorithm that, on every feasible instance, outputs a schedule using at most $\OPT+1$ machines, where $\OPT$ is the true minimum. Unless $\mathrm{P}=\mathrm{NP}$, no polynomial-time algorithm guarantees $\OPT$ machines for any fixed $x\ge2$, so the additive guarantee of one is best possible.