SOLUTION. We will prove first the following result for all real numbers $x_{1}, x_{2}, \ldots, x_{n}$
$$n(n-1) \sum_{i=1}^{n}\left|x_{i}\right|+n|S| \geq \sum_{i, j=1}^{n}\left|x_{i}+(n-1) x_{j}\right|$$
where $S=x_{1}+x_{2}+\ldots+x_{n}$. Indeed, let $z_{i}=\left|x_{i}\right| \forall i \in\{1,2, \ldots, n\}$ and $A=\{i \mid 1 \leq$ $\left.i \leq n, i \in \mathbb{N}, x_{i} \geq 0\right\}, B=\left\{i \mid 1 \leq i \leq n, i \in \mathbb{N}, x_{i}<0\right\}$. WLOG, we may assume that $A=\{1,2, \ldots, k\}$ and $B=\{k+1, k+2, \ldots, n\}$, then $|A|=k,|B|=n-k=m$ and $z_{i} \geq 0$ for all $i \in A \cup B$. The inequality above becomes
$$\begin{array}{r} n(n-1)\left(\sum_{i \in A} z_{i}+\sum_{j \in B} z_{j}\right)+n\left|\sum_{i \in A} z_{i}-\sum_{j \in B} z_{j}\right| \\ \geq \sum_{i, i^{\prime} \in A}\left|z_{i}+(n-1) z_{i^{\prime}}\right|+\sum_{j, j^{\prime} \in B}\left|z_{j}+z_{j^{\prime}}\right|+\sum_{i \in A, j \in B}\left(\left|z_{i}-(n-1) z_{j}\right|+\left|(n-1) z_{i}-z_{j}\right|\right) \end{array}$$

Because $n=k+m$, the previous inequality is equivalent to
$$\begin{aligned} & n(m-1) \sum_{i \in A} z_{i}+n(k-1) \sum_{j \in B} z_{j}+n\left|\sum_{i \in A} z_{i}-\sum_{j \in B} z_{j}\right| \\ \geq & \sum_{i \in A, j \in B}\left|z_{i}-(n-1) z_{j}\right|+\sum_{i \in A, j \in B}\left|(n-1) z_{i}-z_{j}\right| \end{aligned}$$

For each $i \in A$ we denote
$$B_{i}=\left\{j \in B \mid(n-1) z_{i} \geq z_{j}\right\} ; B_{i}^{\prime}=\left\{j \in B \mid z_{i} \geq(n-1) z_{j}\right\}$$

For each $j \in B$ we denote
$$A_{j}=\left\{i \in A \mid(n-1) z_{j} \geq z_{i}\right\} ; A_{j}^{\prime}=\left\{i \in A \mid z_{j} \geq(n-1) z_{i}\right\}$$

We have of course $B_{i}^{\prime} \subset B_{i} \subset B$ and $A_{i}^{\prime} \subset A_{i} \subset A$. After giving up the absolute value signs, the right-hand side expression of $(\star)$ is indeed equal to
$$\sum_{i \in A}\left(m n-2\left|B_{i}^{\prime}\right|-2(n-1)\left|B_{i}\right|\right) z_{i}+\sum_{j \in B}\left(k n-2\left|A_{j}^{\prime}\right|-2(n-1)\left|A_{j}\right|\right) z_{j}$$

WLOG, we may assume that $\sum_{i \in A} z_{i} \geq \sum_{j \in B} z_{j}$. The inequality above becomes
$$\sum_{i \in A}\left(\left|B_{i}^{\prime}\right|+(n-1)\left|B_{i}\right|\right) z_{i}+\sum_{j \in B}\left(\left|A_{j}^{\prime}\right|+(n-1)\left|A_{j}\right|-n\right) z_{j} \geq 0$$

Notice that if for all $j \in B$, we have $\left|A_{j}^{\prime}\right| \geq 1$, then the conclusion follows immediately (because $A_{j}^{\prime} \subset A_{j}$, then $\left|A_{j}\right| \geq 1$ and $\left|A_{j}^{\prime}\right|+(n-1)\left|A_{j}\right|-n \geq 0 \forall j \in B$ ). If not, we may assume that there exists a certain number $r \in B$ for which $\left|A_{r}^{\prime}\right|=0$, and therefore $\left|A_{r}\right|=0$. Because $\left|A_{r}\right|=0$, it follows that $(n-1) z_{r} \leq z_{i}$ for all $i \in A$. This implies that $\left|B_{i}\right| \geq\left|B_{i}^{\prime}\right| \geq 1$ for all $i \in A$, therefore $\left|B_{i}^{\prime}\right|+(n-1)\left|B_{i}\right| \geq n$ and we conclude that
$$\sum_{i \in A}\left(\left|B_{i}^{\prime}\right|+(n-1)\left|B_{i}\right|\right) z_{i}+\sum_{j \in B}\left(\left|A_{j}^{\prime}\right|+(n-1)\left|A_{j}\right|-n\right) z_{j} \geq n \sum_{i \in A} z_{i}-n \sum_{j \in B} z_{j} \geq 0$$

Therefore (1) has been successfully proved and therefore Suranji's inequality follows immediately from Karamata inequality and the Symmetric Majorization Criterion.