I1.2 If the average of 10 distinct positive integers is $2 \alpha$, what is the largest possible value of the largest integer, $\beta$, of the ten integers?
Let the 10 distinct positive integers be $0<x_{1}<x_{2}<\ldots<x_{10}$, in ascending order.
$$
\begin{array}{l}
\frac{x_{1}+x_{2}+\cdots+x_{10}}{10}=2 \times 5=10 \\
x_{1}+x_{2}+\ldots+x_{9}+\beta=100
\end{array}
$$

If $\beta$ is the largest possible, then $x_{1}, x_{2}, \ldots, x_{9}$ must be as small as possible.
The least possible $x_{1}, x_{2}, \ldots, x_{9}$ are $1,2,3, \ldots, 9$.
The largest possible $\beta=100-(1+2+\ldots+9)=100-45=55$