Five. (15 points) Find the largest positive integer $n$ that satisfies the inequality
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{11}\right]+\left[\frac{n}{13}\right]<n
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Five, Proof: For any integers $x, k$, we have
$$
\left[\frac{x}{k}\right] \geqslant \frac{x-k+1}{k} \text {. }
$$

Fact .. By the division algorithm, there exist integers $q, r$ such that $x=k q+r(0 \leqslant r \leqslant k-1)$. Thus,
$$
\begin{array}{l}
{\left[\frac{x}{k}\right]=\left[\frac{k q+r}{k}\right]=q+\left[\frac{r}{k}\right]=q \text {, }} \\
\frac{x-k+1}{k}=q+\frac{r-(k-1)}{k} \leqslant q=\left[\frac{x}{k}\right] \text {. } \\
\end{array}
$$

Let the positive integer $n$ satisfy the inequality
$$
\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{11}\right]+\left[\frac{n}{13}\right]<n \text {. }
$$

Then $n-1 \geqslant\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{11}\right]+\left[\frac{n}{13}\right]$
$$
\begin{array}{l}
\geqslant \frac{n-1}{2}+\frac{n-2}{3}+\frac{n-10}{11}+\frac{n-12}{13} \\
=\frac{859}{858} n-\frac{2573}{858} .
\end{array}
$$

Thus, $n \leqslant 1715$.
Furthermore, when $n=1715$,
$$
\begin{array}{l}
{\left[\frac{1715}{2}\right]+\left[\frac{1715}{3}\right]+\left[\frac{1715}{11}\right]+\left[\frac{1715}{13}\right]} \\
=857+571+155+131=1714<1715 .
\end{array}
$$

Therefore, the largest positive integer $n$ is 1715.