4. The sequence $x_{1}, x_{2}, \cdots, x_{100}$ satisfies the following conditions: For $k=1,2, \cdots, 100, x_{k}$ is less than the sum of the other 99 numbers by $k$. Given that $x_{S 0}=\frac{m}{n}, m, n$ are coprime positive integers. Then $m+n$ equals $(\quad)$.
(A) 50
(B) 100
(C) 165
(D) 173
4.D.

Let $S=x_{1}+x_{2}+\cdots+x_{100}$, then $x_{k}=\left(S-x_{k}\right)-k$, which means $k+2 x_{k}=S$.
Summing over $k$ we get
$$
(1+2+\cdots+100)+2 S=100 S \text {. }
$$

Therefore, $S=\frac{2525}{49}$. Thus, $x_{50}=\frac{S-50}{2}=\frac{75}{98}$.
Hence, $m+n=173$.