jmc

Since the coefficient of $f(x)$ are real, the nonreal roots of $f(x)$ must come in conjugate pairs. Furthermore, the magnitude of a complex number and its conjugate are always equal. If $n$ is the number of magnitudes $|r_i|$ that correspond to nonreal roots, then $f(x)$ has at least $2n$ nonreal roots, which means it has at most $2006-2n$ real roots.

Also, this leaves $1006-n$ magnitudes that correspond to real roots, which means that the number of real roots is at least $1006-n.$ Hence, $$1006-n\le 2006-2n,$$so $n\le 1000.$ Then the number of real roots is at least $1006-n\ge 6.$

The monic polynomial with roots $\pm i,$ $\pm 2i,$ $\dots ,$ $\pm 1000i,$ 1001, 1002, 1003, 1004, 1005, 1006 satisfies the conditions, and has 6 real roots, so the minimum number of real roots is $\boxed{6}.$