11. Let $a_{1}, a_{2}, \cdots, a_{50}, b_{1}, b_{2}, \cdots, b_{50}$ be distinct numbers, such that the following equation
$$
\left|x-a_{1}\right|+\cdots+\left|x-a_{50}\right|=\left|x-b_{1}\right|+\cdots+\left|x-b_{50}\right|
$$

has a finite number of roots. How many roots can there be at most?
Solve 49.
$$
\text { Let } f(x)=\left|x-a_{1}\right|+\cdots+\left|x-a_{50}\right|-\left|x-b_{1}\right|-\cdots-\mid x-
$$
$b_{50}$ 1. Then the original equation is $f(x)=0$. Let $c_{1}<c_{2}<\cdots<c_{100}$ be the elements of the set $\left\{a_{1}, a_{2}, \cdots, a_{50}, b_{1}, b_{2}, \cdots, b_{50}\right\}$ arranged in increasing order. In the 101 intervals $\left(-\infty, c_{1}\right],\left[c_{1}, c_{2}\right], \cdots,\left[c_{99}, c_{100}\right],\left[c_{100}, \infty\right)$, the function $f(x)$ is linear. In the interval $\left(-\infty, c_{1}\right]$, $f(x)=a_{1}+a_{2}+\cdots+a_{50}-b_{1}-b_{2}-\cdots-b_{50}:=$ $m$ (the symbol “: =” means “is defined as”), and in the interval $\left[c_{100}, \infty\right)$, $f(x)=-m$. Since the number of roots of the equation is finite, $m \neq 0$.

We move along the number line from left to right. Initially, the coefficient of $x$ in $f(x)$ is 0. Each time we pass a $c_{i}$, the opening of one of the absolute values in $f(x)$ changes, causing the coefficient of $x$ to change by $\pm 2$ (increase by 2 or decrease by 2). This indicates that the coefficient of $x$ is always even and does not change sign before it becomes 0. Therefore, the coefficient of $x$ in any two adjacent intervals is either both non-negative or both non-positive. Thus, $f(x)$ is either non-increasing or non-decreasing in such an interval union. As a result, if $f(x)=0$ has only a finite number of roots, then it has no more than 1 root in each of the intervals $\left[c_{1}, c_{3}\right], \cdots$, $\left[c_{97}, c_{99}\right],\left[c_{99}, c_{100}\right]$. Additionally, since $f\left(c_{1}\right)$ and $f\left(c_{100}\right)$ have different signs, and $f(x)$ changes sign at each root, $f(x)=0$ has an odd number of roots, and thus no more than 49 roots.
On the other hand, it is not difficult to verify that if
$\left\{a_{1}, a_{2}, \cdots, a_{50}\right\}=\{n \mid n$ is a positive integer not greater than 100,
and $n \equiv 1$ or $0(\bmod 4)\}$,
$$
\left\{b_{1}, b_{2}, \cdots, b_{50}\right\}=\{1,2, \cdots, 100\} \backslash\left\{a_{1}, a_{2}, \cdots, a_{50}\right\},
$$

then the equation $f(x)=0$ has exactly 49 roots.