Solution:

1. Solution (Complete induction). Let the masses of the 1000 coins in grams, arranged in ascending order, be denoted by $m_{1} \leq m_{2} \leq \cdots \leq m_{1000}$. We first prove, by complete induction on $\ell=1, \ldots, 1000$, the following statement.

Lemma. For all $1 \leq \ell \leq 1000$ and every real number $0 \leq x \leq 2+\sum_{i=1}^{\ell} m_{i}$ there exists a (possibly empty) subset $I \subset\{1,2, \ldots, \ell\}$ such that $\sum_{i \in I} m_{i} \in[x-2, x]$ holds.

We have $x_{1} \leq 2000 / 1000=2$, so the base case $\ell=1$ is clear: if $x<2$, we can choose $I=\emptyset$, otherwise $I=\{1\}$ works.

For the induction step $(\ell \rightarrow \ell+1$, where $\ell \leq 999$) we see that for $x \leq \sum_{i=1}^{\ell} m_{i}$ there is nothing to show. Since we can replace any set $I$ by $I \cup\{\ell+1\}$, the claim also holds for $m_{\ell+1} \leq x \leq m_{\ell+1}+\sum_{i=1}^{\ell} m_{i}=\sum_{i=1}^{\ell+1} m_{i}$. Thus the induction step is complete, except when $2+\sum_{i=1}^{\ell} m_{i}<m_{\ell+1}$. In this case it follows that $m_{i}>2+\ell$ for all $\ell+1 \leq i \leq 1000$, which implies the following inequality.

$$2000=\sum_{i=1}^{1000} m_{i}=\sum_{i=1}^{\ell} m_{i}+\sum_{i=\ell+1}^{1000} m_{i}>\ell+(1000-\ell)(2+\ell)=2000+999 \ell-\ell^{2}$$

Hence $\ell>999$, a contradiction, which completes the proof of the lemma.

We now apply the lemma to $\ell=1000$ and $x=1001$. This shows that there exists a subset $I \subset\{1,2, \ldots, 1000\}$ for which $\sum_{i \in I} x_{i} \in[999,1001]$ holds. Since $\sum_{i=1}^{1000} m_{i}=2000$, we also have $\sum_{i \in\{1, \ldots, 1000\} \backslash I} x_{i} \in[999,1001]$, and the division of the coins corresponding to the partition
$$\{1, \ldots, 1000\}=I \cup(\{1, \ldots, 1000\} \backslash I)$$
satisfies the claim.

2. Solution (Greedy algorithm). We use the same notation as in the first solution. The pirate distributes the coins according to the following procedure: he places all coins, in descending order of their masses (i.e., starting with the heaviest coin of mass $m_{1000}$), one after another onto two piles $H_{1}$ and $H_{2}$, placing each coin on whichever pile is lighter at that moment; in case of a tie he chooses either pile arbitrarily.

We claim that neither pile ever weighs more than 1001 grams.

Suppose that after placing the $k$-th coin of mass $m_{1001-k}$, one of the piles is nevertheless heavier than 1001 grams. Before this step, both piles together weighed $\sum_{i=1002-k}^{1000} m_{i}$ grams, and since the coin was placed on the lighter pile, it follows that

$$\begin{aligned} m_{1001-k}+\frac{1}{2} \sum_{i=1002-k}^{1000} m_{i} & >1001 \\ m_{1001-k}+\frac{1}{2} \sum_{i=1}^{1000} m_{i} & >1001+\frac{1}{2} \sum_{i=1}^{1001-k} m_{i} \\ \frac{1}{2} m_{1001-k}+1000 & >1001+\frac{1}{2} \sum_{i=1}^{1000-k} m_{i} \\ m_{1001-k} & >2+\sum_{i=1}^{1000-k} m_{i} \geq 2+1000-k=1002-k \end{aligned}$$

From this it now follows that

$$\begin{aligned} 2000=\sum_{i=1}^{1000} m_{i} & =\left(\sum_{i=1}^{1000-k} m_{i}\right)+\left(\sum_{i=1001-k}^{1000} m_{i}\right) \\ & \geq 1000-k+k \cdot m_{1001-k} \\ & >1000-k+k(1002-k) \end{aligned}$$

so $f(k)=-k^{2}+1001 k-1000<0$, which, because $f(k)=(k-1)(1000-k)$ and $1 \leq k \leq 1000$, is nevertheless a contradiction.

So the claim is true, and even after distributing the lightest coin of mass $m_{1}$, both piles weigh no more than 1001 grams. Since together they weigh 2000 grams, each pile must weigh at least 999 grams, and we have found a distribution as required.

3. Solution (Pigeonhole principle). We again use the same notation as in the first solution. Since $m_{1000} \geq 2>1$, we have $m_{1}+\cdots+m_{999}<1999$, and the 1000 sums
$$0, m_{1}, m_{1}+m_{2}, \ldots, m_{1}+m_{2}+\cdots+m_{999}$$
each lie in exactly one of the 1000 pigeonholes
$$\begin{gathered} \{[0,1) \cup[1000,1001),\} \\ \{[1,2) \cup[1001,1002),\} \\ \ldots, \\ \{[998,999) \cup[1998,1999),\} \\ \{[999,1000)\} \end{gathered}$$
If one of the sums lies in the last pigeonhole $[999,1000)$, we are done, otherwise two sums lie in the same pigeonhole $[k, k+1) \cup[1000+k, 1001+k)$ (for some fixed $0 \leq k \leq 998$). Since by assumption $m_{i} \geq 1$, one of the sums lies in the interval $[k, k+1)$ and the other in $[1000+k, 1001+k)$, so the difference $m_{i+1}+m_{i+2}+\cdots+m_{j}$ of the two sums lies in the interval $(999,1001)$ and thus yields a distribution as desired.