The sixteenth question, find the largest real number $\lambda$, such that the inequality $\sum_{k=1}^{n} x_{k}^{3}\left(x_{k}-x_{k-1}\right) \geq \frac{1}{4}+\frac{\lambda}{n}$ holds for any positive integer $n$ and any real numbers $0=x_{0} \leq x_{2} \leq \ldots \leq x_{n}=1$.

The sixteenth question,
Solution method one, on the one hand, take $x_{k}=\sqrt{\frac{k}{n}}(0 \leq k \leq n)$, then
$$
\begin{array}{l}
\sum_{k=1}^{n} x_{k}^{3}\left(x_{k}-x_{k-1}\right)=\frac{1}{n^{2}}\left[\sum_{k=1}^{n}\left(k^{2}-k \sqrt{k(k-1)}\right)\right] \\
=\frac{1}{n^{2}}\left[\sum _ { k = 1 } ^ { n } \left(k^{2}-\frac{k}{2} \sqrt{\left.\left.(2 k-1)^{2}-1\right)\right]}\right.\right. \\
\left(\frac{1}{4}+\frac{3}{8}\right) x_{1}{ }^{4}, the conclusion holds.
Assume the conclusion holds for $n$. When $n+1$, according to the induction hypothesis:
$$
\sum_{k=1}^{n+1} x_{k}{ }^{3}\left(x_{k}-x_{k-1}\right)>\left(\frac{1}{4}+\frac{3}{8 n}\right) x_{n}{ }^{4}+x_{n+1}{ }^{3}\left(x_{n+1}-x_{n}\right)
$$

Thus, we only need to prove:
$$
\begin{array}{l}
\left(\frac{1}{4}+\frac{3}{8 n}\right) x_{n}{ }^{4}+x_{n+1}{ }^{3}\left(x_{n+1}-x_{n}\right) \geq\left(\frac{1}{4}+\frac{3}{8(n+1)}\right) x_{n+1}{ }^{4} \\
\Leftrightarrow \frac{3(n+1)}{8(n+1)} x_{n+1}{ }^{4}+\frac{2 n+3}{8 n} x_{n}{ }^{4} \geq x_{n+1}{ }^{3} x_{n}
\end{array}
$$

In fact, according to the mean inequality:
$$
\frac{3(2 n+1)}{8(n+1)} x_{n+1}{ }^{4}+\frac{2 n+3}{8 n} x_{n}{ }^{4} \geq 4 \sqrt[4]{\left(\frac{2 n+1}{8(n+1)} x_{n+1}{ }^{4}\right)^{3} \cdot \frac{2 n+3}{8 n} x_{n}{ }^{4}}>x_{n+1}{ }^{3} x_{n}
$$

Thus (1) holds, and the proposition is proved!
Method two: Let $A=\sum_{k=1}^{n} x_{k}^{3}\left(x_{k}-x_{k-1}\right), B=\sum_{k=1}^{n} x_{k-1}^{3}\left(x_{k}-x_{k-1}\right)$, then
$$
\begin{array}{l}
A+B=\sum_{k=1}^{n}\left(x_{k}^{3}+x_{k-1}{ }^{3}\right)\left(x_{k}-x_{k-1}\right) \\
\geq \frac{\left.\sum_{k=1}^{n}\left(x_{k}{ }^{3}+x_{k} x_{k-1}+x_{k} x_{k-1}+x_{k-1}\right)^{3}\right)\left(x_{k}-x_{k-1}\right)}{2} \\
=\frac{\left.\sum_{k=1}^{n}\left(x_{k}^{4}-x_{k-1}\right)^{4}\right)}{2}=\frac{1}{2} \\
A-B=\sum_{k=1}^{n}\left(x_{k}^{3}-x_{k-1}{ }^{3}\right)\left(x_{k}-x_{k-1}\right) \\
\geq \sum_{k=1}^{n}\left(x_{k}{ }^{2}+x_{k} x_{k-1}+x_{k-1}\right)\left(x_{k}-x_{k-1}\right)^{2} \\
=\frac{3 \sum_{k=1}^{n}\left(x_{k}+x_{k-1}\right)^{2}\left(x_{k}-x_{k-1}\right)^{2}}{4}=\frac{3 \sum_{k=1}^{n}\left(x_{k}{ }^{2}-x_{k-1}\right)^{2}}{4} \\
\geq \frac{3\left(\sum_{k=1}^{n}\left(x_{k}{ }^{2}-x_{k-1}^{2}\right)\right)^{2}}{4 n}=\frac{3}{4 n}
\end{array}
$$

From (2)+(3), we know $A \geq \frac{1}{4}+\frac{3}{8 \mathrm{n}}$, and the proposition is proved!