Argentine National Olympiad 2015

Label the unit segments $S_1,S_2,S_3,\dots$ in the order they appear on $S$ from left to right. Suppose that $S_k$ and $S_{k+2}$ have a common point for some $k$. Then their union is a longer segment that contains $S_{k+1}$. So the latter can be removed and $S$ will still be completely covered, contrary to the hypothesis. It follows that $S_k$ and $S_{k+2}$ have no points in common (not even one). In particular the odd-indexed segments $S_1,S_3,S_5,\dots$ are disjoint, with segments of positive length separating $S_{2i-1}$ and $S_{2i+1}$ for each $i$. There can be at most $49$ such unit segments on a segment of length $50$. This implies that there are at most $98$ segments in our covering system. Indeed if there were at least $99$ of them then at least $50$ would be odd-indexed, which is impossible.

Consider the interval $I=[0,50]$ on the numerical line. Mark on it the terms of the arithmetic progression with first term $\frac{1}{2}$, last term $\frac{99}{2}$ and length $98$; its common difference is $d=\frac{49}{97}>\frac{1}{2}$.

Place $98$ unit segments $S_1,S_2,\dots ,S_{98}$ on $I$ so that their midpoints coincide with the terms of the progression. It is clear that they cover $I$ completely; also $S_1$ and $S_{98}$ are the only segments containing $0$ and $50$ respectively. Consider $S_k$ and $S_{k+2}$ where $1\le k\le 96$. Their midpoints are at distance $2d=\frac{98}{97}>1$, hence they are separated by a gap of length $\frac{1}{97}$. The gap is covered only by segment $S_{k+1}$. It follows that no segment $S_k$ with $2\le k\le 97$ can be removed without spoiling the covering. By the same reason neither is it possible to remove $S_1$ or $S_{98}$. So $S_1,S_2,\dots ,S_{98}$ is a covering system with the desired properties. It has a maximum number of segments, equal to $98$, which is the answer to our question.