jmc

Since the inequality is always true for $M=0,$ it suffices to consider the case $M\ne 0.$

For a particular $c$ and for any tuple $(x_1,\dots ,x_{101})$ satisfying the conditions, the tuple $(-x_1,\dots ,-x_{101})$ satisfies the conditions as well, so we may assume that $M>0.$ Finally, we may assume that $x_1\le x_2\le \cdots \le x_{101},$ so that $M=x_{51}.$

We want to find the largest $c$ such that the inequality $$x_1^2+x_2^2+\cdots +x_{101}^2\ge c{x}_{51}^2$$always holds, where $x_1\le x_2\le \cdots \le x_{101}$ and $x_1+x_2+\cdots +x_{101}=0.$ Therefore, fixing a value of $x_{51},$ we should write inequalities that minimize $x_1^2+x_2^2+\cdots +x_{101}^2.$

To compare the terms on the left-hand side to $x_{51}^2,$ we deal with the terms $x_1^2+x_2^2+\cdots +x_{50}^2$ and $x_{51}^2+x_{52}^2+\cdots +x_{101}^2$ separately.

By Cauchy-Schwarz, $$(1+1+\cdots +1)(x_1^2+x_2^2+\cdots +x_{50}^2)\ge (x_1+x_2+\cdots +x_{50}{)}^2,$$so $$x_1^2+x_2^2+\cdots +x_{50}^2\ge \frac{1}{50}{\left(x_1+x_2+\cdots +x_{50}\right)}^2.$$We have $x_1+x_2+\cdots +x_{50}=-x_{51}-x_{52}-\cdots -x_{101}\le -51x_{51}$ because $x_{51}\le x_{52}\le \cdots \le x_{101}.$ Since $x_{51}>0,$ both $x_1+x_2+\cdots +x_{50}$ and $-51x_{51}$ are negative, so we can write $$\begin{array}{rl}{x}_1^2+x_2^2+\cdots +x_{50}^2& \ge \frac{1}{50}(x_1+x_2+\cdots +x_{50}{)}^2\\ & \ge \frac{1}{50}{\left(-51x_{51}\right)}^2\\ & =\frac{51^2}{50}{x}_{51}^2.\end{array}$$On the other hand, since $0<x_{51}\le x_{52}\le \cdots \le x_{101},$ we simply have $$x_{51}^2+x_{52}^2+\cdots +x_{101}^2\ge 51x_{51}^2.$$Putting all this together gives $$(x_1^2+x_2^2+\cdots +x_{50}{)}^2+(x_{51}^2+x_{52}^2+\cdots +x_{101}^2)\ge \left(\frac{51^2}{50}+51\right)x_{51}^2=\frac{5151}{50}{x}_{51}^2.$$Equality holds when $x_1=x_2=\cdots =x_{50}=-\frac{51}{50}$ and $x_{51}=x_{52}=\cdots =x_{101}=1,$ so the answer is $\boxed{\frac{5151}{50}}.$