1926 * Given $a_{1}=2.56, a_{2}=2.61, a_{3}=2.65, a_{4}=2.71, a_{5}=2.79$, $a_{6}=2.82, a_{7}=2.86$ these 7 numbers sum up to 19. Now replace each $a_{i}$ with an integer $A_{i}$, such that $A_{1}+A_{2}+\cdots+A_{7}=19$, and the "error" $\left|A_{i}-a_{i}\right|$ has the smallest possible maximum value $M$. For the smallest $M, 100 M=$ ?
Analysis According to the problem, $A_{i}$ can only be 2 or 3. Since $\sum_{i=1}^{7} A_{i}=19$, it follows that among $A_{i}(i=1,2, \cdots, 7)$, 5 take the value 3 and 2 take the value 2. Noting that $a_{1}, a_{2}$ are closest to 2, we set $A_{1}=A_{2}=2, A_{3}=A_{4}=\cdots=$ $A_{7}=3$. $\left((2.61-2)+\left(3-a_{i}\right)<(3-2.61)+\left(a_{i}-2\right)\right.$, so $A_{1}=2$. Similarly, $A_{2}=$ 2). Thus, $M=\max _{1 \leq i<}\left|A_{i}-a_{i}\right|=0.61$, hence $100 M=61$.