11. Given an integer array consisting of 121 integers, where each number is between 1 and 1000 (repetition is allowed), this array has a unique mode (i.e., the integer that appears most frequently). Let $D$ be the difference between this mode and the arithmetic mean of the array. When $D$ reaches its maximum value, what is $[D]$ equal to?
11. Let $M$ and $m$ be the "mode" and "arithmetic mean" respectively. Without loss of generality, assume $M \geqslant m$. To maximize $D$, then $M=1000$ (otherwise, if $M=1000-k$, then increasing $M$ by $k$ while $m$ increases by no more than $k$, $D$ is obviously non-decreasing). In the case of $M=1000$, we must make $m$ as small as possible. At this time, 1000 must appear at least twice in the array. If 1000 appears exactly twice, then the other numbers can only appear once. In this case, when the other (121-2) numbers are $1,2, \cdots, 119$,

If $M=1000$ appears 3 times, then the other numbers can appear at most 2 times. At this time, when the other 118 numbers are $1,1,2,2, \cdots, 59,59$, $m$ is the smallest, $m=\frac{\frac{118(59+1)}{2}+3000}{121}=\frac{6540}{121}$. If $M$ appears 4 times, then the smallest value of $m$ is $\frac{\frac{117(39+1)}{2}+4000}{121}=\frac{6340}{121}$. If $M$ appears 5 times, the smallest value of $m$ is $\frac{\frac{116(29+1)}{2}+5000}{121}=\frac{6740}{121}$. If $M$ appears 6 times, then the smallest value of $m$ is $\frac{115(23+1)+6000}{121}=\frac{7380}{121}$. If $M=1000$ 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 the smallest, $m=\frac{6340}{121}=52+\frac{48}{121}$. The maximum value of $D$ is $1000-(52+\frac{48}{121})$, so $[D]=947$.