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
$$\begin{aligned} I_k &= [k-1, k), \quad 1 \le k < M \\ I_M &= [M-1, L]. \end{aligned}$$
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 $a$
$$J_k = [(k-1)(1+a), (k-1)(1+a)+a], \quad 1 \le k \le M.$$
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 $y-x > k(1+a) - ((k-1)(1+a)+a) = 1$.