= Solution
For $n\geq2$, the <probability integral transform> makes $U_i=F(X_i)$ iid uniform on $[0,1]$. Continuity of $F$ is enough; strict increase is unnecessary. Their extreme <order statistics> have joint density
$$
f_{U_{(1)},U_{(n)}}(u,v)=n(n-1)(v-u)^{n-2},\qquad0<u<v<1:
$$
choose the observations giving the two extremes and put the remaining $n-2$ between them. Transform to $r=v-u$ and integrate $u$ from zero to $1-r$. The <uniformized sample range> therefore has
$$
\boxed{f_R(r)=n(n-1)r^{n-2}(1-r),\quad F_R(r)=nr^{n-1}-(n-1)r^n,\quad0<r<1.}
$$
Equivalently $R\sim\operatorname{Beta}(n-1,2)$. For $n=1$, the range is identically zero instead.
The exact failure probability is
$$
\mathbb P(R<1-\varepsilon)=(1-\varepsilon)^{n-1}[1+(n-1)\varepsilon].
$$
Thus the exact integer sample size is the least $n\geq2$ making this quantity at most $\delta$. For $\varepsilon\downarrow0$ with $n\varepsilon\to\vartheta>0$, it tends to $(1+\vartheta)e^{-\vartheta}$. This function decreases strictly from one to zero on positive $\vartheta$, because its derivative is $-\vartheta e^{-\vartheta}$. For $0<\delta<1$ there is therefore exactly one positive solution of
$$
(1+\vartheta)e^{-\vartheta}=\delta,\qquad\text{equivalently}\quad1+\vartheta-\delta e^\vartheta=0.
$$
This gives the <extreme-order-statistic tolerance interval> sample-size rule
$$
\boxed{n\approx\vartheta/\varepsilon.}
$$
When $\delta$ is small, $\vartheta=\log(1/\delta)+\log(1+\vartheta)$, so a useful first refinement is $\vartheta\approx\log(1/\delta)+\log\log(1/\delta)$. This is an approximation, not an exact integer guarantee; use the preceding exact failure probability to check rounding and very stringent tail requirements. The interval here encloses population content, rather than estimating a fixed distribution parameter.
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