Solution:

Answer: $503$, $1509$

Let $x$ denote the number Tom chooses. By the symmetry of the problem, picking $x$ and picking $2012-x$ yield the same expected profit. If Tom picks $1006$, Dick sees that if he picks $1007$, Harry's best play is to pick $1005$, and Dick will win with probability $\frac{1005}{2011}$, and clearly this is the best outcome he can achieve. So Dick will pick $1007$ (or $1005$) and Harry will pick $1005$ (or $1007$), and Tom will win with probability $\frac{1}{2011}$. Picking the number $2$ will make the probability of winning at least $\frac{2}{2011}$ since Dick and Harry would both be foolish to pick $1$, so picking $1006$ is suboptimal. It is now clear that the answer to the problem is some pair $(x, 2012-x)$.

By the symmetry of the problem we can assume without loss of generality that $1 \leq x \leq 1005$. We will show that of these choices, $x=503$ maximizes Tom's expected profit. The trick is to examine the relationship between Tom's choice and Dick's choice. We claim that (a) if $x>503$, Dick's choice is $2013-x$ and (b) if $x<503$, Dick's choice is $1341+\left\lfloor\frac{x+1}{3}\right\rfloor$. Let $y$ denote the number Dick chooses.

Proof (a). Note first that if $x>503$, $2013-x$ is the smallest value of $y$ for which Harry always chooses a number less than $x$. This is obvious, for if $y<2011-x$, a choice of $2012-x$ will give Harry a greater expected profit than a choice of any number less than $x$, and if $y=2012-x$, Harry never chooses a number between $x$ and $y$ since $x-1>\frac{2012-x-x}{2}=1006-x$ holds for all integers $x$ greater than $503$, so, by symmetry, Harry chooses a number less than $x$ exactly half the time. The desired result actually follows immediately because the fact that Harry always chooses a number less than $x$ when $y=2013-x$ implies that there would be no point in Dick choosing a number greater than $2013-x$, so it suffices to compare Dick's expected profit when $y=2013-x$ with that of all smaller values of $y$, which is trivial.

Proof (b). This case is somewhat more difficult. First, it should be obvious that if $x<503$, Harry never chooses a number less than $x$. The proof is by contradiction. If $y \leq 2011-x$, Harry can obtain greater expected profit by choosing $2012-x$ than by choosing any number less than $x$. If $y \geq 1002+x$, Harry can obtain greater expected profit by choosing any number between $x$ and $y$ than by choosing a number less than $x$. Hence if Harry chooses a number less than $x$, $x$ and $y$ must satisfy $y>2011-x$ and $y<1002+x$, which implies $2x>2009$, contradiction. We next claim that if $y \geq 1342+\left\lfloor\frac{x+1}{3}\right\rfloor$, then Harry chooses a number between $x$ and $y$. The proof follows from the inequality $\frac{1342+\left\lfloor\frac{x+1}{3}\right\rfloor-x}{2}>2011-\left(1342+\left\lfloor\frac{x+1}{3}\right\rfloor\right)$, which is equivalent to $3\left\lfloor\frac{x+1}{3}\right\rfloor>x-4$, which is obviously true. We also claim that if $y \leq 1340+\left\lfloor\frac{x+1}{3}\right\rfloor$ then Harry chooses a number greater than $y$. The proof follows from the inequality $\frac{1340+\left\lfloor\frac{x+1}{3}\right\rfloor-x}{2}<2011-\left(1340+\left\lfloor\frac{x+1}{3}\right\rfloor\right)$, which is equivalent to $3\left\lfloor\frac{x+1}{3}\right\rfloor<x+2$, which is also obviously true. Now if $x \not\equiv 1 \pmod{3}$, then we have $3\left\lfloor\frac{x+1}{3}\right\rfloor>x-1$, so, in fact, if $y=1341+\left\lfloor\frac{x+1}{3}\right\rfloor$, Harry still chooses a number between $x$ and $y$. In this case it is clear that such a choice of $y$ maximizes Dick's expected profit. It turns out that the same holds true even if $x \equiv 1 \pmod{3}$; the computations are only slightly more involved. We omit them here because they bear tangential relation to the main proof.

We may conclude that if $x>503$, the optimal choice for $x$ is $504$, and if $x<503$, the optimal choice for $x$ is $502$. In the first case, $y=1509$, and, in the second case, $y=1508$. Since a choice of $x=503$ and $y=1509$ clearly outperforms both of these combinations when evaluated based on Tom's expected profit, it suffices now to show that if Tom chooses $503$, Dick chooses $1509$. Since the computations are routine and almost identical to those shown above, the proof is left as an exercise to the reader.