Example 6 Given an integer array consisting of 121 integers, each integer in this array takes a value between 1 and 1000 (inclusive of 1 and 1000), and repeated values are allowed. The arithmetic mean of these numbers is $m$, and there is a unique "mode" (the number that appears most frequently) $M$ in this set of numbers. Let $D=M-m$. If $D$ is as large as possible, find $[D]$ (where $[D]$ represents the greatest integer not exceeding $D$).
Obviously, to make $D$ as large as possible, the mode $M$ should be as large as possible. For this purpose, let $M=1000$, and the mean $m$ should be as small as possible, so the other numbers should be as small as possible, i.e., $1,2,3, \cdots$.

Thus, the key lies in the frequency of the mode $M$, i.e., how many times $M$ appears for $D$ to be maximized.

Let $M=1000$ appear $x$ times, then $1,2, \cdots$, $\frac{121-x}{x-1}$ each appear $x-1$ times (when $\frac{121-x}{x-1}$ is an integer). At this point, the arithmetic mean is
$$
\begin{aligned}
m & =\frac{(x-1)\left[1+2+\cdots+\frac{121-x}{x-1}\right]+1000 x}{121} \\
& =\frac{20}{121} \cdot \frac{50 x^{2}-53 x+363}{x-1} .
\end{aligned}
$$

Let $m_{1}=\frac{50 x^{2}-53 x+363}{x-1}$, we now find the minimum value of $m_{1}$. The equation
$$
50 x^{2}-\left(53+m_{1}\right) x+121 \times 3+m_{1}=0
$$

has real roots, then
$$
\begin{array}{l}
\Delta=\left(53+m_{1}\right)^{2}-4 \times 50\left(121 \times 3+m_{1}\right) \\
\quad \geqslant 0, \\
m_{1}^{2}-94 m_{1}-69791 \geqslant 0 .
\end{array}
$$

Thus, $m_{1} \geqslant 315$, $m \geqslant \frac{20}{121} \times 315=52 \frac{8}{121}$, i.e., $m>52$.
At this point, $D=1000-m<948$.
Therefore, $[D]=947$. At this point, $x=4$ can be obtained.
Thus, a set of integers with $[D]=947$ can be constructed:
$$
1,1,1,2,2,2, \cdots, 39,39,39,1000,1000 \text {, }
$$
$1000,1000$.