60 Let real numbers $a_{1}, \cdots, a_{n}$ satisfy $a_{1}<a_{2}<\cdots<a_{n}$. When $x$ is a real number, try to express the minimum value of $\sum_{k=1}^{n}\left|x-a_{k}\right|$ in terms of $a_{1}, a_{2}, \cdots, a_{n}$.
60 Let $f(x)=\sum_{k=1}^{n}\left|x-a_{k}\right|$.
For any two real numbers $a, b$, it is clear that
$$
|x-a|+|x-b| \geqslant|a-b|
$$

and the equality holds if and only if $x$ is in the interval $[a, b]$.
Thus, when $n=2 m$,
$$
\left|x-a_{m+1-k}\right|+\left|x-a_{m+k}\right| \geqslant a_{m+k}-a_{m+1-k}, k=1,2, \cdots, m .
$$

Therefore,
$$
f(x) \geqslant \sum_{k=1}^{m} a_{m-k}-\sum_{k=1}^{m} a_{k} .
$$

The equality holds when $x \in\left[a_{m}, a_{m+1}\right]$.
When $n=2 m+1$, similarly,
$$
f(x) \geqslant \sum_{k=2}^{m+1} a_{m+k}-\sum_{k=1}^{m} a_{k} .
$$

The equality holds when $x=a_{m+1}$.