$\begingroup$

Most of these answers take an empirical approach; I would rather take an analytical approach, although I'm not the most qualified to do so.

There are two kinds of numbers that could be good choices:

Large numbers that share common factors with many smaller numbers, and

Small numbers that share common factors with few larger numbers

Large Numbers

The best large numbers will be the product of as many primes as possible, but to be in range all of them should be less than $\sqrt{max}$. The more primes you use the smaller they must be, and smaller primes have more multiples.

The obvious first-guess here is the largest Primorial within range, but for the range 2 through 1000 that guess is only 210(=2×3×5×7), which is too small. The largest multiple in range is 840.

A number chosen in this way will win against half of lower numbers by the common factor 2, against $\frac{1}{3}$ of the other half by the common factor 3, against $\frac{1}{5}$ of the remaining $\frac{1}{3}$ by the common factor 5, and so on. It will lose against the same proportion of higher numbers.

840 will win against about $\frac{27}{35}$ of the 838 lower numbers and about $\frac{8}{35}$ of the 159 higher numbers, for a win rate of about $\frac{683}{998}\approx 68.3\%$.

I'm not sure the 7 is helping, so I'm also going to look at the next lower primorial, 30(=2×3×5); the largest multiple in range is 990, which will win against $\frac{11}{15}$ of the 989 lower numbers and $\frac{4}{15}$ of the 10 higher numbers, for a win rate of about $\frac{730}{998}\approx 73.1\%$

So it looks like being higher is more valuable than being more composite - but 1000, a multiple of 2 and 5, only wins against $\frac{3}{5}$ of the 998 lower numbers (for a win rate of about 59.9%), so there must be a limit to that.

The largest multiple of 6(=2×3) in range is 996, with a win rate of about 66.4%, so it looks like the limit on the value of compositeness in this case is between 5# and 7#.

Small Numbers

In order to have few common factors with larger numbers, the best small numbers will be (relatively) small primes, near $\sqrt{max}$. Any prime will lose against all smaller numbers, but will only lose against larger numbers which are multiples of itself. If $\exists p_n = \sqrt{max}$, that prime will lose against $p_n-2$ lower numbers and $p_n-1$ higher numbers.

Moving down from $\sqrt{max}$ loses fewer below and more above, and moving up loses move below and fewer above. Exactly how much these change depends on the gaps between primes, which I don't know how to consider in general, but the difference in wins and losses can be expressed in terms of the gap.

Choose some $p_n = \sqrt{max}-gap$; $p_n$ will lose to $p_n-2$ lower numbers and $\lfloor\frac{max}{p_n}\rfloor-1$ higher numbers. Losses are minimized where $p_n+\lfloor\frac{max}{p_n}\rfloor$ is minimized. If $p_n$ could be chosen continuously, that would be at $p_n=\sqrt{max}$; since it cannot be chose continuously, it is the $p_n$ nearest to $\sqrt{max}$ and any neighboring $p_n$ which lose equally.

In this case $\sqrt{1000}\approx 31.6$, so the nearest primes are 31, 29, 37, and 23, in that order.

31($\approx\sqrt{max}$-0.6) will lose to 29 lower numbers and $\lfloor \frac{1000}{31} \rfloor-1 = 31$ higher numbers, for a win rate of $\frac{938}{998}\approx 94.0\%$

29($\approx\sqrt{max}$-2.6) will lose to 27 lower numbers and $\lfloor\frac{1000}{29}\rfloor-1 = 33$ higher numbers, for a win rate of $\frac{938}{998}\approx 94.0\%$

37($\approx\sqrt{max}$+5.4) will lose to 35 lower numbers and $\lfloor\frac{1000}{37}\rfloor-1 = 26$ higher numbers, for a win rate of $\frac{937}{998}\approx 93.9\%$

23($\approx\sqrt{max}$-8.6) will lose to 21 lower numbers and $\lfloor\frac{1000}{23}\rfloor-1 = 42$ higher numbers, for a win rate of $\frac{935}{998}\approx 93.7\%$

The two primes closest to $\sqrt{1000}$ are a tie for the winningest choice.