 Solution

LetIgnoring the immaterial choice of density values at the sample points, the likelihood function isBetween consecutive order statistics, $N(\theta )$ is constant while $(1+1/\theta {)}^{N(\theta )}$ decreases with $\theta$. A maximum can therefore be moved to the left endpoint of one of these intervals. With an equivalent version using $1_{\{x\le \theta \}}$, the maximum is attained at a sample order statistic. Thus the maximum-likelihood estimator coincides with one of $X_1,\dots ,X_n$.