2. The sum of the following 7 numbers is exactly 19:
$$
\begin{array}{l}
a_{1}=2.56, a_{2}=2.61, a_{3}=2.65, a_{4}=2.71, a_{5}=2.79, a_{6}= \\
2.82, a_{7}=2.86 .
\end{array}
$$

We want to use integers $A_{i}$ to approximate $a_{i}$ $(1 \leqslant i \leqslant 7)$, such that the sum of $A_{i}$ is still 19, and the maximum value $M$ of the error $|A_{i}-a_{i}|$ is as small as possible. Then, for this smallest $M$, what is $100 M$?
[Solution] Since $21$.
To make $A_{1}+A_{2}+\cdots+A_{7}=19$, then $A_{1}, A_{2}, \cdots, A_{7}$ should consist of five 3s and two 2s.

To make $M$ as small as possible, the smallest two numbers $a_{1}$ and $a_{2}$ should have their approximations as $A_{1}=A_{2}=2$, and the others should be $A_{3}=A_{4}=A_{5}=A_{6}=A_{7}=3$. In this case,
$$
\begin{array}{l}
M=a_{2}-A_{2}=2.61-2=0.61, \\
100 M=61 .
\end{array}
$$