Answer: 947

Small Hint:

By symmetry, place the mode at the low endpoint and push every other entry as high as the frequency restriction allows

Big Hint:

If the mode occurs $f$ times, every other value may occur at most $f-1$ times; optimize separately over $f$

Solution:

By reflecting every value $x$ to $1001-x,$ it suffices to maximize the mean minus the mode. For a fixed modal frequency $f,$ the extremal sample has $f$ copies of $1,$ then fills the largest available integers with at most $f-1$ copies each.

Put $121-f=q(f-1)+r,$ where $0\leq r<f-1.$ The nonmodal entries are $f-1$ copies of each of $1000,$ $999,$ $\ldots,$ $1001-q,$ followed by $r$ copies of $1000-q.$ For $f=2,$ $f=3,$ $f=4,$ $f=5,$ and $f=6,$ this formula gives floors $924,$ $945,$ $947,$ $944,$ and $939,$ respectively. If $f\geq7,$ there are at most $114$ nonmodal terms, so even the weaker bound $D\leq\frac{114(999)}{121}<942$ suffices. Thus the maximum occurs at $f=4.$

The extremal sample contains four $1$’s and three copies of every integer from $962$ through $1000.$ Put $T=962+963+\cdots+1000.$ Since $T=38259,$ $$\begin{aligned}D&=\frac{3T+4}{121}-1\\&=\frac{114660}{121}\\&=947+\frac{73}{121}.\end{aligned}$$ Therefore the largest possible floor is $947.$