(Notation: $\sum_{(a)}^{} ! {y_i}$ = sum over all permutations (of the degrees) of $y_1^{a_1}$$y_2^{a_2}$$...$$y_n^{a_n}$; sequences $(a)$=($a_1,a_2,...,a_n$), and $(y)$=($y_1,y_2,...,y_n$). $a)$ Let $x_1=-y_1, x_2=-y_2, ..., x_n=-y_n$, i.e. $x_i=-y_i$ $\forall$ 1 $\le$ $i$ $\le$ $n$. Therefore, since $x_i$ $\in$ $\mathbb{R^-}$ $\Rightarrow$ $y_i$ $\in$ $\mathbb{R^+}$. Hence $-m$=$\sum_{i=1}^{n} -x_i$=$\sum_{i=1}^{n} y_i$, and $S$=$\sum_{i=1}^{n} (x_i\sum_{j=i+1}^{n} x_j)$=$\sum_{i=1}^{n} (-y_i\sum_{j=i+1}^{n} -y_j)$=$\sum_{i=1}^{n} ((-1)(-1)y_i\sum_{j=i+1}^{n} y_j)$=$\sum_{i=1}^{n}(y_i\sum_{j=i+1}^{n} y_j)$. Now let us look at the sequences $(a)$ and $(b)$, both having $n$-terms: $(a)$=$(2,0,0,0,0,$...$,0,0)$, $(b)$=$(1,1,0,0,0,$...$,0,0)$. We see that $(a)$$\succ$$(b)$, and from Muirhead's inequality over $y_i$ and sequences $(a)$ and $(b)$ we get $\sum_{(a)}^{} ! {y_i}$ $\ge$ $\sum_{(b)}^{} ! {y_i}$. Namely, $(n-1)!$($y_1^{2}$ + $y_2^{2}$ + ... + $y_n^{2}$)$\ge$$(n-2)!$($y_1$$y_2$ + $y_1$$y_3$ + $...$ + $y_1$$y_n$ + $y_2$$y_3$ + $y_2$$y_4$ + $...$ + $y_2$$y_n$ + $...$ + $y_{n-1}$$y_n$)=$(n-2)!$$S$. From here we get $S$$\le$$(n-1)$$\sum_{i}^{} {y_i}^{2}$. Therefore, the maximum value for $S$ is $S_{max}$=$(n-1)$$\sum_{i}^{} {y_i}^{2}$ when equality is satisfied for the last inequality. Since $(a)$$\not\equiv$$(b)$, equality is satisfied when $y_1$ = $y_2$ = $...$ = $y_n$ = $Y$, and hence $x_1$ = $x_2$ = $...$ = $x_n$ = $X$. Now, $Y$=$\frac{-m}{n}$, $X$=$\frac{m}{n}$. Hence, $S_{max}$=$(n-1)nX^2$, and ${S_{max}=\frac{(n-1)m^2}{n}}$. Therefore, the maximum value of $S$ is when all $x_i$ are equal, namely $\boxed{S_{max}=\frac{(n-1)m^2}{n}}$. $b)$ Through analogy as in case $a)$ (we will be working with $x_i$ instead of $y_i$ since $x_i$ $\in$ $\mathbb{R^+}$), the maximum value of $S$ is when all $x_i$ are equal, namely $\boxed{S_{max}=\frac{(n-1)m^2}{n}}$.