Let $\xi=(\xi_{1},\ldots,\xi_{N})\in\mathbb{R}^{N}$ and $\varepsilon >0$, I need to solve the following optimization problem:
$$ \begin{array}{cl} {\displaystyle \sup_{x\in\mathbb{R}^{N}} } & {\displaystyle \frac{1}{N}\sum_{i=1}^{N}(\xi_{i}-x_{i}) } \\ \text{subject to} & {\displaystyle \frac{1}{N}\sum_{i=1}^{N}\left|x_{i}\right| \leq \varepsilon } \end{array} $$
Remark: In this link they deal with optimization problems with inequality constraints, my problem is that in my case the constraints involved absolute values and I do not see how this information can help.
Since $\xi$ is constant, the problem is equivalent to $$\min \frac{1}{N}\sum_{i=1}^N x_i, \text{ s.t. } \frac{1}{N}\sum_{i=1}^N |x_i|\leq \epsilon.$$ Note that if we change $x_i$ by $-x_i,$ te constraint is not affected. Hence we can restrict our attention to the case when $x_i\leq 0.$ But then we have
$$\frac{1}{N}\sum_{i=1}^N -x_i=\frac{1}N \sum_{i=1}^N |x_i|\leq \epsilon,$$ from which we find
$$\frac 1 N \sum_{i=1}^N x_i\geq -\epsilon.$$ It is now easy to see that $\bar{x}=-\epsilon e$ is a solution $(e=(1,\ldots,1)^T)$.