14. The function $f(x)$ defined on the interval $[1,2017]$ satisfies $f(1)=f(2017)$, and for any $x, y \in [1,2017]$, it holds that $|f(x)-f(y)| \leqslant 2|x-y|$. If the real number $m$ satisfies that for any $x, y \in [1,2017]$, it holds that $|f(x)-f(y)| \leqslant m$, find the minimum value of $m$.
14. First, prove: For any $x, y \in [1,2017]$, we have $|f(x)-f(y)| \leqslant 2016$.
In fact, if $|x-y| \leqslant 1008$, then $|f(x)-f(y)| \leqslant 2|x-y| \leqslant 2016$;
if $|x-y| > 1008$, assume without loss of generality that $1 \leqslant x < y \leqslant 2017$, then
$$
\begin{array}{l}
|f(x)-f(y)| \\
=|f(x)-f(1)+f(2017)-f(y)| \\
\leqslant|f(x)-f(1)|+|f(2017)-f(y)| \\
\leqslant 2|x-1|+2|2017-y| \\
\leqslant 2(x-1)+2(2017-y) \\
=2 \times 2016-2(y-x) \\
<2 \times 2016-2 \times 1008=2016 .
\end{array}
$$

Next, prove that 2016 is the smallest.
Consider the function $f(x)=2|x-1009|$, then the function $f(x)$ satisfies
$$
f(1)=f(2017)=2016,
$$

and for any $x, y (1 \leqslant x < y \leqslant 2017)$, we have
$$
\begin{array}{l}
|f(x)-f(y)| \\
=2|| x-1009|-| y-1009|| \\
\leqslant 2|x-y| \leqslant 2016 .
\end{array}
$$

Taking $x=1, y=1009$, we get
$$
|f(1)-f(1009)|=|2016-0|=2016.
$$

Therefore, 2016 is the smallest.
In conclusion, the smallest value of the real number $m$ is 2016.