$$
\begin{array}{l}
\text { 4. If } x_{1}>x_{2}>x_{3}>x_{4}>0, \text { and the inequality } \\
\log _{\frac{x_{1}}{x_{2}}} 2014+\log _{\frac{x_{2}}{x_{3}}} 2014+\log _{\frac{x_{3}}{x_{4}}} 2014 \\
\geqslant k \log _{\frac{x_{1}}{}} 2014
\end{array}
$$

always holds, then the maximum value of the real number $k$ is $\qquad$
4.9.

From the given, we have
$$
\begin{array}{l}
\frac{\ln 2014}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln 2014}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln 2014}{\ln \frac{x_{3}}{x_{4}}} \\
\geqslant k \frac{\ln 2014}{\ln \frac{x_{1}}{x_{4}}} .
\end{array}
$$

Since $x_{1}>x_{2}>x_{3}>x_{4}>0$, we have
$$
\begin{array}{l}
\ln \frac{x_{1}}{x_{2}}>0, \ln \frac{x_{2}}{x_{3}}>0, \ln \frac{x_{3}}{x_{4}}>0, \\
\ln \frac{x_{1}}{x_{4}}=\ln \frac{x_{1}}{x_{2}}+\ln \frac{x_{2}}{x_{3}}+\ln \frac{x_{3}}{x_{4}}>0 . \\
\text { Therefore, } k \leqslant \frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{3}}{x_{4}}} . \\
\text { And } \frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{3}}{x_{4}}} \\
=\left(\ln \frac{x_{1}}{x_{2}}+\ln \frac{x_{2}}{x_{3}}+\ln \frac{x_{3}}{x_{4}}\right) \text {. } \\
\left(\frac{1}{\ln \frac{x_{1}}{x_{2}}}+\frac{1}{\ln \frac{x_{2}}{x_{3}}}+\frac{1}{\ln \frac{x_{3}}{x_{4}}}\right) \\
\geqslant 9 \text {, } \\
\end{array}
$$

Equality holds if and only if $\ln \frac{x_{1}}{x_{2}}=\ln \frac{x_{2}}{x_{3}}=\ln \frac{x_{3}}{x_{4}}$, i.e., $\frac{x_{1}}{x_{2}}=\frac{x_{2}}{x_{3}}=\frac{x_{3}}{x_{4}}$.
Therefore, the maximum value of the real number $k$ is 9.