I must be true. Note that
$\begin{array}{rcl}
m_1 & = & \dfrac{x_1+x_2+\ldots+x_{100}}{100} \\
100m_1 & = & x_1+x_2+\ldots+x_{100}
\end{array}$
Then, we have
$\begin{array}{rcl}
m_2 & = & \dfrac{x_1+x_2+\ldots+x_{100}+m_1}{101} \\
& = & \dfrac{100m_1+m_1}{101} \\
& = & m_1
\end{array}$
II must be true. Note that the mean is greater than the smallest data and smaller than the greatest data. Therefore, The greatest and smallest values of the two sets of data remain unchanged. Hence, $r_1=r_2$.
III must be false. Note that
$\begin{array}{rcl}
v_1 & = & \dfrac{(x_1-m_1)^2+\ldots+(x_{100}-m_1)^2}{100}
\end{array}$
Then, we have
$\begin{array}{rcl}
v_2 & = & \dfrac{1}{101}\left[(x_1-m_2)^2+\ldots+(x_{100}-m_2)^2+(m_1-m_2)^2\right] \\
& = & \dfrac{(x_1-m_1)^2+\ldots+(x_{100}-m_1)^2}{101} \\
& < & \dfrac{(x_1-m_1)^2+\ldots+(x_{100}-m_1)^2}{100} \\
& = & v_1
\end{array}$
