= Solution
A statistic $T$ is <sufficient statistic>[sufficient] for $\theta$ if the conditional distribution of the full sample given $T$ does not depend on $\theta$. It is <minimal sufficient statistic>[minimal sufficient] if it is a function of every sufficient statistic.
The likelihood factors as
$$
L(\theta;\mathbf x)=(2\theta)^{-n}\mathbf1_{\{M\le2\theta\}},
$$
so the <Fisher-Neyman factorization theorem> shows that $M$, and hence its one-to-one transform $\widehat\theta=M/2$, is sufficient. Moreover, for two samples $\mathbf x$ and $\mathbf y$, the ratio $L(\theta;\mathbf x)/L(\theta;\mathbf y)$ is independent of $\theta$ exactly when their maxima agree. The likelihood-ratio criterion for minimal sufficiency therefore shows that $M$ and $\widehat\theta$ are minimal sufficient.
For $n\ge2$, the <sample mean> $\widetilde\theta$ is not sufficient: two samples can have the same mean but different maxima, and their likelihood ratio then depends on $\theta$ through the support indicators. It is consequently not minimal sufficient. For the degenerate special case $n=1$, the sample mean and maximum coincide and both conclusions reverse.
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