IRL_ABooklet

Let $M=\lfloor L\rfloor$ be the least integer greater than or equal to $L$. We claim that the least value of $N$ is $9M+1$.

First, we show that $9M+1$ points are enough. Partition $[0,L]$ into subintervals $I_1,\dots ,I_M$, where $$\begin{array}{rl}{I}_k& =[k-1,k),1\le k<M\\ I_M& =[M-1,L].\end{array}$$ These intervals are of length at most $1$ and, given at least $9M+1$ distinct points in $[0,L]$, the average number of points in each of them is at least $(9M+1)/M>9$, and so one such subinterval must contain strictly more than $9$ points $x_i$, as required.

For the converse direction, we first note that $a:=(L-(M-1))/M$ is positive. Define the subintervals of length $a$ $$J_k=[(k-1)(1+a),(k-1)(1+a)+a],1\le k\le M.$$ Since $$(M-1)(1+a)+a=M-1+Ma=L,$$ every $J_k$ is a subinterval of $[0,M]$. By picking nine distinct points in each $J_k$, we claim that we have a set of $9M$ points such that there are not $10$ of these points all within a distance $1$ of each other. To justify the claim, it suffices to note that if $x\in J_k$ and $y\in J_{k+1}$ then $y-x>k(1+a)-((k-1)(1+a)+a)=1$.