Four, find the largest real number $C$ such that the inequality
$$
\sum_{k=1}^{100} \frac{1}{x_{k}} \sum_{k=1}^{100}\left(x_{k}^{3}+2 x_{k}\right)-\left(\sum_{k=1}^{100} \sqrt{x_{k}^{2}+1}\right)^{2} \geqslant C
$$

holds for any 100 distinct positive integers $x_{1}, x_{2}, \cdots, x_{100}$.
(Supplied by Wenzhou High School)
Let $m=100$.
From $\left(\sum_{k=1}^{m} \sqrt{x_{k}^{2}+1}\right)^{2}$, we think of the Cauchy-Schwarz inequality and the Lagrange identity. Consider transforming the left side of the original expression into $S+T$, where,
$$
\begin{aligned}
S & =\sum_{k=1}^{m} \frac{1}{x_{k}} \sum_{k=1}^{m} x_{k}, \\
T & =\sum_{k=1}^{m} \frac{1}{x_{k}} \sum_{k=1}^{m} x_{k}\left(x_{k}^{2}+1\right)-\left(\sum_{k=1}^{m} \sqrt{x_{k}^{2}+1}\right)^{2} \\
& =\sum_{1 \leqslant i < j \leqslant m} \frac{1}{x_{i} x_{j}}\left(x_{i} \sqrt{x_{i}^{2}+1}-x_{j} \sqrt{x_{j}^{2}+1}\right)\left(x_{j} \sqrt{x_{j}^{2}+1}-x_{i} \sqrt{x_{i}^{2}+1}\right) \\
& =\sum_{1 \leqslant i < j \leqslant m} A\left(x_{j}, x_{i}\right)\left(x_{j}-x_{i}\right),
\end{aligned}
$$
where
$$
\begin{array}{l}
A(y, x)=\sqrt{\frac{y\left(y^{2}+1\right)}{x}}-\sqrt{\frac{x\left(x^{2}+1\right)}{y}} \\
=\frac{y \sqrt{y^{2}+1}-x \sqrt{x^{2}+1}}{\sqrt{x y}} .
\end{array}
$$

It is easy to see that $x \sqrt{x^{2}+1}$ is close to $x^{2}+\frac{1}{2}$, and the latter is slightly greater than the former, because
$$
\begin{array}{l}
g(x) \triangleq x^{2}+\frac{1}{2}-x \sqrt{x^{2}+1} \\
=\frac{1}{4} \times \frac{1}{x^{2}+\frac{1}{2}+x \sqrt{x^{2}+1}}\frac{(y+x)(y-x)}{\sqrt{x y}}>2(y-x), \\
A(y, x)=\frac{y^{2}-x^{2}}{\sqrt{x y}}+\frac{g(x)-g(y)}{\sqrt{x y}} \\
\end{array}
$$
for $i < j$, we have $x_{j}-x_{i} \geqslant j-i$.
$$
\begin{array}{l}
\text { Then } T=\sum_{1 \leqslant i < j \leqslant m} A\left(x_{j}, x_{i}\right)\left(x_{j}-x_{i}\right) \\
\geqslant \sum_{1 \leqslant i < j \leqslant m} 2\left(j-i\right) \\
=\frac{1}{3}\left(m^{4}-m^{2}\right),
\end{array}
$$
and when
$$
\left(x_{1}, x_{2}, \cdots, x_{m}\right)=(N+1, N+2, \cdots, N+m),
$$

and the positive integer $N$ is sufficiently large, we have
$$
A\left(x_{j}, x_{i}\right) \approx 2,
$$
so
$$
T \approx \sum_{1 \leqslant i < j \leqslant m} 2\left(j-i\right) = \frac{1}{3} m^{2}\left(m^{2}+2\right).
$$

When $\left(x_{1}, x_{2}, \cdots, x_{m}\right)=(N+1, N+2, \cdots, N+m)$ and the positive integer $N$ is sufficiently large, $S+T$ is sufficiently close to its minimum value.
Therefore, the desired value is $\frac{1}{3} m^{2}\left(m^{2}+2\right)=33340000$.