= Solution
Write $L=\min_iX_i$, $R=\max_iX_i$ and $M=\max_i|X_i|=\max\{-L,R\}$. The joint <probability density> factors as
$$
f_\theta(x)=(2\theta)^{-n}\mathbf1_{\{\theta\geq\max_i|x_i|\}}=(2\theta)^{-n}\mathbf1_{\{\theta\geq\max(-L,R)\}}.
$$
By the <Fisher-Neyman factorization theorem>, \b[$T=(L,R)$ is a <sufficient statistic>]. The same factorization shows that the smaller <statistic> $M$ is also a <sufficient statistic>. This gives the <sufficient statistic for a symmetric uniform sample>.
In fact, the <likelihood-ratio criterion for minimal sufficiency> identifies $M$ as a <minimal sufficient statistic>: two sample points with the same $M$ have identical likelihoods for all $\theta$; with different $M$, their likelihoods have different parameter supports, and so are not proportional as functions of $\theta$. Equivalently, the condition that $f_\theta(x)=c f_\theta(y)$ for every $\theta$ and some positive constant $c$ holds exactly when their absolute maxima agree.
To see rigorously that $T$ cannot be recovered from $M$ almost surely, fix any $\theta>0$. The sample has a <distribution> invariant under simultaneous sign reversal. If $T=h(M)$ almost surely, the same identity would hold for the sign-reversed sample on another probability-one set. Since $M$ is unchanged by sign reversal, it would follow that
$$
(L,R)=(-R,-L)\quad\text{almost surely},
$$
and hence $L=-R$ almost surely. But this event has probability zero for a finite sample from a continuous <uniform distribution>: it entails $X_i=-X_j$ for some pair of indices, including the possible case $X_i=0$. Therefore $T$ is not a function of the sufficient statistic $M$. \b[The pair of extrema is not minimally sufficient], for every $n\geq1$, even though it is sufficient.
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