Every possible length

Let $\ell_i$ be the length of $I_i.$ Let $D_{ij} = \{ x-y : x \in I_i, y \in I_j\}.$ Then it should be clear that $D_{ij}$ is a closed interval inside $[-1,1]$ of length $\ell_i + \ell_j.$

The condition on distances implies that the union of the $D_{ij},$ for $1\le i,j \le 10,$ is the entire interval $[-1,1].$ So then $\begin{aligned} 2 \le \sum\limits_{i=1}^{10} \sum\limits_{j=1}^{10} \text{length}(D_{ij}) &= \sum\limits_{i=1}^{10} \sum\limits_{j=1}^{10} (\ell_i + \ell_j) \\ &= \sum\limits_{i=1}^{10} (10\ell_i + s) \\ &= 10s + 10s = 20s. \end{aligned}$ So we get $s \ge 1/10.$

Is this in fact the minimum? Well, the inequality is an equality if and only if the length of the overlap between any two $D_{ij}$ is zero. But if the lengths of $I_i$ and $I_j$ are nonzero, it's immediately clear that $D_{ii}$ and $D_{jj}$ have an overlap of nonzero length. So equality can hold only if at most one of the intervals has nonzero length. And in fact, $S = [0,1/10] \cup \{2/10\} \cup \{3/10\} \cup \cdots \cup \{9/10\} \cup \{1\}$ has the all-distances property, with $s=1/10.$ So the answer is $1+10 = \fbox{11}.$

This argument shows, I believe, that the only sets where the minimum is attained are the one above and its mirror image (nine points plus $[9/10,1]$ ).

Also, there was nothing special about $10$ ; the same argument shows that if $S$ consists of $k$ disjoint intervals, the minimum possible value of the sum of the lengths of the intervals is $1/k.$