11.1. Let $a_{1}, a_{2}, \cdots, a_{50}, b_{1}, b_{2}, \cdots, b_{50}$ be distinct numbers such that the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{50}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{50}\right|
\end{array}
$$

has a finite number of roots. How many roots can there be at most?
11.1.49.
$$
\begin{array}{l}
\text { Let } f(x)=\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{50}\right|- \\
\left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{50}\right| \text {. } \\
\end{array}
$$

Thus, 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 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 each of the 101 intervals, and in the interval $\left(-\infty, c_{1}\right]$, $f(x)=a_{1}+a_{2}+\cdots+a_{50}-b_{1}-b_{2}-\cdots-b_{50}=m$, and in the interval $\left[c_{100},+\infty\right)$, $f(x)=-m$.
Since the number of roots of the equation is finite, $m \neq 0$.
Next, move along the number line from left to right. Initially, the coefficient of $x$ in $f(x)$ is 0. Each time a $c_{i}$ is crossed, the opening method 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 a union of intervals. As a result, if $f(x)=0$ has only a finite number of roots, then it has at most one 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, if
$$
\begin{array}{l}
\left\{a_{1}, a_{2}, \cdots, a_{50}\right\} \\
=\{1,4,5,8,9,12, \cdots, 89,92,93,96,98,99\} \text {, } \\
\left\{b_{1}, b_{2}, \cdots, b_{50}\right\} \\
=\{2,3,6,7,10,11, \cdots, 90,91,94,95,97,101\} \text {, } \\
\text { then } f(1)=-1, f(2)=f(3)=1 \text {, } \\
f(4)=f(5)=-1, f(6)=f(7)=1, \cdots \text {, } \\
f(92)=f(93)=-1, f(94)=f(95)=1 \text {, } \\
f(96)=f(97)=-1, f(98)=f(99)=-3 \text {, } \\
f(101)=1 \text {. } \\
\end{array}
$$

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