Original answer
This question when applied to the Earth is purely academic. The easiest solution is the one posted by Johannes: One revolution per one hour and 24 minutes. Why go beyond that? The question is academic.
It's not academic when applied to asteroids. There's an interesting effect, the Yarkovsky–O'Keefe–Radzievskii–Paddack effect, aka the YORP effect, that can make asteroids spin faster and faster and faster. This theoretical spin-up was observed in the asteroid formerly named as 2000 PH5 (astronomers tend to have obscure naming conventions). After confirming that this asteroid was spinning faster than before, it was renamed as 54509 YORP.
An asteroid that spins faster than the gravitational attraction needed to hold it together is likely to split in two. This is now hypothesized to be the reason we see so many binary asteroids. See Walsh, Richardson, and Michel, "Rotational breakup as the origin of small binary asteroids," Nature 454.7201 (2008): 188-191 for details.
Updated answer: Going beyond Johannes's answer
Johannes's answer assumed a spherical Earth. That's not that a particularly good assumption; that the Earth is rotating creates an equatorial bulge. This bulge will grow ever larger as the Earth's rotation rate increases. This means the rotation rate needed to feel weightless at the equator will be substantially less than one revolution per 5070 seconds.
That the Earth has an equatorial bulge is a consequence of the second law of thermodynamics. The Earth's surface is an equipotential surface where potential energy is computed from the perspective an Earth-fixed frame (a frame rotating with the Earth). Anything but this would violate the principle of minimum energy.
Using a spherical harmonic expansion of the Earth's gravity field and ignoring higher moments, the rotating frame potential energy is given by (e.g., see http://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-201-essentials-of-geophysics-fall-2004/lecture-notes/ch2.pdf equation 2.41)
$$U(r,\lambda) = -\left(\frac{\mu}{r} \left(1 - J_2 \frac {a^2} {r^2} \left(\frac 3 2 \sin^2 \lambda - \frac 1 2\right) \right) + \frac 1 2 r^2\omega^2 \cos^2 \lambda\right)$$
where $\mu \equiv GM_\text{earth} = 398600.4418\,\text{km}^3/\text{s}^2$ is the Earth's geopotential, $J_2$ is a widely-used measure of the Earth's oblateness, $\lambda$ is the geocentric latitude, $r$ is the radius of the Earth at that latitude, $a$ is the equatorial radius of the Earth, and $\omega$ is the Earth's rotation rate.
The perceived gravitational acceleration is the negated gradient of this potential:
$$g(r,\lambda) = -\nabla U(r,\lambda) = -\left(\frac{\partial U}{\partial r} \hat r + \frac 1 r \frac{\partial U}{\partial \lambda} \hat \lambda\right)$$
At the equator, the non-radial component vanishes and the radial component becomes
$$g_e =
\frac{\mu}{a^2} \left(1 + \frac 3 2 \, J_2\right) - a \omega^2$$
We want to make this zero (this is the condition that makes things weightless at the equator). Thus
$$\omega^2 = \frac{\mu}{a^3} \left(1 + \frac 3 2 \, J_2\right)$$
The potential needs to be the same at all points on the surface of the Earth per the principle of minimum energy. In particular, the potential at the poles and the equator need to be the same:
$$
\frac \mu c\left (1-\frac {a^2}{c^2} J_2\right) =
\frac \mu a \left(1+\frac 1 2 J_2\right) + \frac 1 2 a^2\omega ^2
$$
where $c$ is the polar radius of the Earth. Using the value of $\omega$ that results in weightless at the equation, this becomes$$
\frac \mu c\left (1-\frac {a^2}{c^2} J_2\right) =
\frac \mu a \left(1+\frac 1 2 J_2\right) + \frac 1 2 a^2 \frac{\mu}{a^3} \left(1 + \frac 3 2 \, J_2\right) =
\frac \mu a \left(\frac 3 2 + \frac 5 4 J_2\right)
$$ or
$$ \frac 1 c\left (1-\frac {a^2}{c^2} J_2\right) = \frac 1 a \left(\frac 3 2 + \frac 5 4 J_2\right)$$
Solving for $\kappa \equiv \frac c a$ yields
$$\kappa^3 \left(\frac 3 2 + \frac 5 4 J_2\right) -\kappa^2 + J_2 = 0$$
So, apparently a cubic. It's worse than that. $J_2$ is a function of flattening and hence is not a constant. This document suggests using $J_2 = \frac 2 3 f - \frac{a^3 \omega^2}{3\mu}$. As a sanity check, this expression yields a value of 0.00108141, which agrees rather nicely with the established value of $J_2$, 0.00108263. The flattening $f$ is given by $f=\frac{a-c} a = 1-\frac c a = 1 -\kappa$, and from above, we have $\frac {a^3 \omega^2}{\mu} = 1+\frac 3 2 J_2$. Solving for $J_2$ yields $J_2 = \frac 2 9 (1-2\kappa)$. Substituting this in the expression for $\kappa$ yields the quartic
$$5 \kappa^4 - 16 \kappa^3 +9\kappa^2 + 4 \kappa - 2 = 0$$
This quartic has four real zeros, three of which ($\kappa\approx-0.469$, $\kappa=1$, and $\kappa\approx 2.30$) can be rejected as non-physical. This leaves but one solution, $\kappa\approx 0.3709102193$. In other words, we have an Earth that is considerably flattened.
Applying $J_2 = \frac 2 9 (1-2\kappa)$ to the expression $\omega^2 = \frac{\mu}{a^3} \left(1 + \frac 3 2 \, J_2\right)$ for the Earth's angular velocity yields
$$\omega^2 = \frac{\mu}{a^3} \left(1 + \frac 3 2 \, \frac 2 9 (1-2\kappa)\right) = \frac 2 3 \frac {\mu}{a^3} (2-\kappa)$$
We now know $\kappa$, but we don't yet know $a$, the Earth's equatorial radius. One way around this problem is to assume that the volume of the Earth is conserved, even in the face of this extreme flattening. This yields $a^3\kappa^2 = {a_0}^3$, where $a_0$ is the Earth's volumetric radius, 6371.0008 km. With this result, the angular velocity that results in objects at the surface of the Earth at the equator being weightless is given by
$$\omega^2 = \frac 2 3 \frac {\mu}{{a_0}^3} \kappa^2(2-\kappa)$$
from whence the desired rotation period is about 3 hours and 38 minutes.
This is considerably greater than the value suggested by assuming a spherical Earth.
