II. (40 points) Let $k \geqslant 1$. Find the largest real number $\lambda$ such that for any real numbers $x_{i}>0$ $(i=1,2, \cdots, n)$ satisfying $\sum_{i=1}^{n} \frac{1}{x_{i}}=2016$, we have
$$
\lambda \sum_{i=1}^{n} \frac{x_{i}}{1+x_{i}} \leqslant\left[\sum_{i=1}^{n} \frac{1}{x_{i}^{k}\left(1+x_{i}\right)}\right]\left(\sum_{i=1}^{n} x_{i}^{k}\right) .
$$
Let $x_{1}=x_{2}=\cdots=x_{n}=\frac{n}{2016}$, then $\lambda \leqslant 2016$.
Below is the proof:
$$
2016 \sum_{i=1}^{n} \frac{x_{i}}{1+x_{i}} \leqslant\left[\sum_{i=1}^{n} \frac{1}{x_{i}^{k}\left(1+x_{i}\right)}\right]\left(\sum_{i=1}^{n} x_{i}^{k}\right) \text {. }
$$

Replace $\frac{1}{x_{i}}$ with $x_{i}$, then equation (1) is equivalent to
$$
\left(\sum_{i=1}^{n} \frac{x_{i}^{k+1}}{1+x_{i}}\right)\left(\sum_{i=1}^{n} \frac{1}{x_{i}^{k}}\right)-\left(\sum_{i=1}^{n} \frac{1}{1+x_{i}}\right)\left(\sum_{i=1}^{n} x_{i}\right) \geqslant 0 \text {. }
$$

The left side of the above equation
$$
\begin{array}{l}
=\sum_{i \neq j}\left(\frac{x_{i}^{k+1}}{1+x_{i}} \cdot \frac{1}{x_{j}^{k}}\right)-\sum_{i \neq j} \frac{x_{j}}{1+x_{i}}=\sum_{i \neq j} \frac{x_{i}^{k+1}-x_{j}^{k+1}}{\left(1+x_{i}\right) x_{j}^{k}} \\
=\frac{1}{2} \sum_{i \neq j}\left[\frac{x_{i}^{k+1}-x_{j}^{k+1}}{\left(1+x_{i}\right) x_{j}^{k}}+\frac{x_{j}^{k+1}-x_{i}^{k+1}}{\left(1+x_{j}\right) x_{i}^{k}}\right] \\
=\frac{1}{2} \sum_{i \neq j}\left(x_{i}^{k+1}-x_{j}^{k+1}\right) \frac{\left(1+x_{j}\right) x_{i}^{k}-\left(1+x_{i}\right) x_{j}^{k}}{\left(1+x_{i}\right)\left(1+x_{j}\right) x_{i}^{k} x_{j}^{k}} \\
=\frac{1}{2} \sum_{i \neq j}\left(x_{i}^{k+1}-x_{j}^{k+1}\right) \frac{\left(x_{i}^{k}-x_{j}^{k}\right)+x_{i} x_{j}\left(x_{i}^{k-1}-x_{j}^{k-1}\right)}{\left(1+x_{i}\right)\left(1+x_{j}\right) x_{i}^{k} x_{j}^{k}} \\
\geqslant 0 .
\end{array}
$$

Therefore, inequality (1) holds.
Thus, $\lambda_{\max }=2016$.