Solution:
Let us define a $k$-configuration to be a finite set of points on the plane such that for each point $P$ in the set, there are at least $k$ points of the set 1 inch from $P$. Then the problem is asking us to show that there is a 2000-configuration with $3^{1000}$ points.

Notice that an equilateral triangle is a 2-configuration which has 3 points. Now, let $A$ be a $k$-configuration with $N$ points, and let $T$ be an equilateral triangle with unit side length. We shall show that it is possible to "add" $A$ and $T$ to create a $(k+2)$-configuration with $3N$ points:

Define the set $S$ by
$$S = \{ a + t \mid a \in A,\ t \in T \}$$
where we treat the points as vectors. In other words, $S$ consists of the vector sums of every point in $A$ with every point in $T$. Since $A$ has $N$ points, and $T$ has 3 points, the set $S$ will have $3N$ points as long as all of these sums are distinct. For the time being, let us assume that the sums are all distinct.

Now we will show that $S$ is a $(k+2)$-configuration. Consider any point $a + t$ in $S$, where $a \in A$ and $t \in T$. Since $A$ is a $k$-configuration, there are $k$ points $a_1, \ldots, a_k \in A$ that are 1 inch away from $a$. Likewise, there are two points $t_1, t_2$ in $T$ which are each 1 inch away from $t$. It is easy to check that the $k+2$ points
$$t + a_1, \ldots, t + a_k ;\ a + t_1, a + t_2$$
are each 1 inch away from $a + t$. Thus $S$ is a $(k+2)$-configuration.

But how do we ensure that all $3N$ sums are distinct? The sums fail to be distinct only if there are pairs $a, a' \in A$ and $t, t' \in T$ with $a + t = a' + t'$, which in turn is true if and only if $a - a' = t' - t$. To ensure that this does not happen, it suffices to rotate one of the two sets (say, $T$) so that the slopes of all of the lines connecting all pairs of points in $T$ do not equal any of the slopes in $A$ (easy to do since there are finitely many points).

For example, in the following diagram, we attempt to "add" two equilateral triangles (the second triangle is outlined), but because of equal slopes, the sum contains only 6 points.

On the other hand, if we rotate the second triangle (in this case, by 30 degrees), the resulting sum contains 9 points (and you should check that this new set is indeed a 4-configuration).

Clearly, we can continue this summation process, adding additional copies of equilateral triangles (making sure to rotate so that no slopes are equal). For each triangle that we add, the new set will have three times as many points. Thus if we add 1000 triangles, we will get a set with $3^{1000}$ points which is a $2 + 2 + \cdots + 2 = 2000$-configuration.