16. Given an array of 121 integers, where each number is between 1 and 1000 (repetition allowed), this set of numbers has a unique mode (i.e., the number that appears most frequently). Let $D$ be the difference between this mode and the arithmetic mean of this set of numbers. What is $[D]$ when $D$ reaches its maximum value?
16. Let $M$ and $m$ be the "remainder" and "arithmetic mean" respectively. Without loss of generality, assume $M \geqslant m$. To maximize $D$, we need $M=1000$ (otherwise, if $M=1000-k$, then increasing $M$ by $k$ while $m$ increases by no more than $k$ would not decrease $D$). If $M=1000$ appears exactly twice, then the other numbers in the set can only appear once. To minimize $m$, the other $(121-2)$ numbers are $1, 2, 3, \cdots, 119$, in which case $m=\frac{9140}{121}$. If $M=1000$ appears 3 times, the other numbers can appear at most twice, and when the other 118 numbers are $1, 1, 2, 2, \cdots, 59, 59$, $m$ is minimized, $m=\frac{6540}{121}$. Similarly, if 1000 appears 4 times, the smallest $m=\frac{6340}{121}$; if 1000 appears 5 times, the smallest $m=\frac{6740}{121}$; if 1000 appears 6 times, the smallest $m=\frac{7380}{121}$; if 1000 appears $n$ times $(n \geqslant 7)$, then $m \geqslant \frac{1000 n}{121} \geqslant \frac{7000}{121}$. Therefore, when $M=1000$ appears exactly 4 times, $m$ is minimized, $m=\frac{6340}{121}=52+\frac{48}{121}$. The maximum value of $D$ is $1000-\left(52+\frac{48}{121}\right), [D]=947$.