A [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every [integer](https://artofproblemsolving.com/wiki/index.php/Integer) $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$.
Let the sum of all of the terms in the sequence be $\mathbb{S}$. Then for each integer $k$, $x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k$. Summing this up for all $k$ from $1, 2, \ldots, 100$,
\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ 100\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\ \mathbb{S}&=\frac{2525}{49}\end{align*}
Now, substituting for $x_{50}$, we get $2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}$, and the answer is $75+98=\boxed{173}$.