Choosing an arithmetic set $\{u, v, w\}$, we can assume that $u < v < w$ holds, and we can therefore write $u = a - d$, $v = a$ and $w = a + d$ with $d > 0$. We now wish this set to also be harmonic. If some number $q$ is the harmonic mean of numbers $p$ and $r$, we have $\frac{1}{p} + \frac{1}{r} = \frac{2}{q}$, which is equivalent to $qr - 2rp + pq = 0$. If $v$ is the harmonic mean of $u$ and $w$, this equation yields $a(a + d) - 2(a + d)a + (a - d)a = 2d^2 = 0$, which means that all three elements of the set are equal, which is a contradiction. There can therefore be no such subsets. If $w$ is the harmonic mean of $u$ and $v$, we obtain
$$(a+d)a - 2a(a-d) + (a-d)(a+d) = 3ad - d^2 = d(3a-d) = 0,$$
which yields $d = 3a$, since $d = 0$ is not possible. We see that any set of the form $\{-2a, a, 4a\}$ has the required properties. Finally, if $u$ is the harmonic mean of $v$ and $w$, we obtain $(a-d)(a+d) - 2(a+d)a + a(a-d) = -3ad - d^2 = -d(3a+d) = 0$, which yields $d = -3a$ and therefore the same sets as the previous case. Since $a$ can assume any integer value not equal to $0$ such that $-2011 < 4a < 2011$, we have $-502 \le a \le 502$, and we see that there are $1004$ subsets with the required properties.