1. The numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy the following condition: For $k=1,2, \cdots, 100, a_{k}$ is less than the sum of the other 99 numbers by $k$. Then $a_{50}=$
$$
\text { II. 1. } \frac{75}{98} \text {. }
$$

Let $S=a_{1}+a_{2}+\cdots+a_{10}$. Then
$$
a_{k}=\left(S-a_{k}\right)-k \Rightarrow k+2 a_{k}=S \text {. }
$$

Summing over $k$ we get
$$
\begin{array}{l}
(1+2+\cdots+100)+2 S=100 S \\
\Rightarrow S=\frac{2525}{49} \Rightarrow a_{50}=\frac{S-50}{2}=\frac{75}{98} .
\end{array}
$$