HMMT, 2014

Solution:
Answer: $\frac{5151}{50}$ OR 103.02 OR $103 \frac{1}{50}$

Suppose without loss of generality that $x_{1} \leq \cdots \leq x_{101}$ and $M = x_{51} \geq 0$.

Note that $f(t) = t^{2}$ is a convex function over the reals, so we may "smooth" to the case $x_{1} = \cdots = x_{50} \leq x_{51} = \cdots = x_{101}$ (the $x_{51} = \cdots$ is why we needed to assume $x_{51} \geq 0$). Indeed, by Jensen's inequality, the map $x_{1}, x_{2}, \ldots, x_{50} \rightarrow \frac{x_{1}+\cdots+x_{50}}{50}, \ldots, \frac{x_{1}+\cdots+x_{50}}{50}$ will decrease or fix the LHS, while preserving the ordering condition and the zero-sum condition.

Similarly, we may without loss of generality replace $x_{51}, \ldots, x_{101}$ with their average (which will decrease or fix the LHS, but also either fix or increase the RHS). But this simplified problem has $x_{1} = \cdots = x_{50} = -51 r$ and $x_{51} = \cdots = x_{101} = 50 r$ for some $r \geq 0$, and by homogeneity, $C$ works if and only if
$$C \leq \frac{50(51)^{2} + 51(50)^{2}}{50^{2}} = \frac{51(101)}{50} = \frac{5151}{50}.$$