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Using Cauchy-Schwarz inequality, we get $$\begin{gathered} x_1+x_2+\cdots +x_n\le |x_1|+|x_2|+\cdots +|x_n| \\ \le \sqrt{n(x_1^2+x_2^2+\cdots +x_n^2)}. \end{gathered}$$

a) Without loss of generality, we can assume $a\le b\le c$ and the inequality turns out to $$\begin{gathered} |a-b|+|b-c|+|c-a|=(a-b)+(b-c)+(c-a) \\ =2(a-c)\le 2\sqrt{2(a^2+c^2)}\le 2\sqrt{2}. \end{gathered}$$ The equality holds when $(a,b,c)=\left(-\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2}\right)$.

b) Without loss of generality, suppose that $a_1$ is the smallest number. We investigate the following cases:

If the sequence $a_1,a_2,\dots ,a_{2019}$ is not decreasing then $S$ can be simplified as $$S=2|a_1-a_{2019}|\le 2\sqrt{2(a_1^2+a_{2019}^2)}\le 2\sqrt{2}.$$ Otherwise, there must exist $k$ such that $1<k<2019$, $a_k\ge a_{k-1}$ and $a_k\le a_{k+1}$. Then $|a_k-a_{k-1}|+|a_{k+1}-a_k|=|a_{k+1}-a_{k-1}|$. Hence, we only need to consider 2018 real numbers and can restate this problem as follows: Given 2018 real numbers $b_1,b_2,\dots ,b_{2018}$ that satisfy $b_1^2+b_2^2+\cdots +b_{2018}^2\le 1$. Find the maximum value of $$S=|b_1-b_2|+|b_2-b_3|+\cdots +|b_{2018}-b_1|.$$ We obtain that $$\begin{gathered} S\le (|b_1|+|b_2|)+(|b_2|+|b_3|)+\cdots +(|b_{2018}|+|b_1|) \\ =2\sum _{1\le i\le 2018}|b_i|\le 2\sqrt{2018}. \end{gathered}$$ The equality holds when all absolute values are equal and the adjacent numbers have different signs. Note that 2018 is even then we choose $b_1=-b_2=b_3=-b_4=\cdots =-b_{2018}=\frac{\sqrt{2018}}{2018}$.

Back to the original problem, we also get $\max S=2\sqrt{2018}$ and the equality holds in many cases, such as $$a_1=-a_2=a_3=-a_4=\cdots =-a_{2018}=\frac{\sqrt{2018}}{2018},a_{2019}=0.$$