9 pirates have to divide 1000 coins...

In order for pirates to make consistent decisions, we must start with two assumptions. First, pirates are all completely logical, and that they are all completely logical is 'common knowledge' (see joriki's comment below). Second, we must take a stance on the bloodthirsty vs. friendly issue (see Micah's comment above), and the problem isn't as interesting with friendly pirates, so we assume they are bloodthirsty.

Under these assumptions, it can be shown that with a loot of $G$ gold coins, and $n$ pirates, the first pirate has a proposal that will be accepted in which he receives $G-\lfloor\frac{n-1}{2}\rfloor$ gold coins (here we assume $G\ge\lfloor\frac{n-1}{2}\rfloor$). Denoting a proposal as $(g_1,\ldots,g_n)$, where pirate $i$ receives $g_i$ coins, the first pirate should propose $(G-\lfloor\frac{n-1}{2}\rfloor,0,1,0,\ldots,\frac{1-(-1)^n}{2}).$

Immediately we note that in any proposal, exactly $\lfloor\frac{n}{2}\rfloor$ pirates receive no gold coins.

We prove that this proposal will be accepted by induction. If $n=1$, the pirate will choose the $G$ gold coins over suicide, and if $n=2$, the first pirate will vote for his own proposal, thus obtaining a majority of the vote.

Next, suppose the said proposal is accepted for $n-1$, and assume there are $n$ pirates. Since all pirates are completely logical, there are exactly $\lfloor\frac{n-1}{2}\rfloor$ pirates that know they will receive $0$ coins if they don't accept the proposal of the first pirate. Since pirates are bloodthirsty, the first pirate must buy their votes with $1$ gold coin each. This gives a proposal that passes with a majority $\lfloor\frac{n-1}{2}\rfloor+1$ votes, where the extra vote is from the first pirate himself.

In the case of $9$ pirates and $1000$ gold coins, the first pirate will receive $996$ gold coins.

HINT: Start with 2 pirates. Distribute money so as to ensure max profit. Then go on adding 1 pirate at a time and figure out the distributions. It is a classic game theory question.