Can an evacuated metal sphere be made such that it can float in the air? Is it possible to construct an evacuated metal sphere which can float in the air without imploding? 
 A: I doubt if a vacuum metal sphere can be constructed that is light enough to have an overall density less than that of air. One of the lightest and strongest metals is beryllium ($d=1.85$) but since as the sphere has to be thin enough to save weight, it's unlikely to be able to withstand the outside pressure of $1\:\mathrm{bar}$.
But a large, thin beryllium sphere filled with hydrogen (at atmospheric pressure) could certainly be made to float in air, if you get the dimensions right.
A: The answer to "is it possible to build?" is "Not with current technology/materials".
However, in theory, this is of course possible.
Some years ago I did the math to reach an expression that relates thickness of a sphere shell and the density of its material, with the minimum diameter that would allow it to float in air. However, I tried some finite element method simulations with some well known materials, and the conclusion was clearly no (at least considering a spheric shell).
I can post the equation later when get home, but it's pretty straightforward to get there. At the time, I remember I tried considering metallic foams (such as aluminum foam) with a density of ~0.3 g/cm3, with a thickness of only a few centimeters, and the minimum diameter was something between 50 and 100 meters.
A: It is not possible with a homogeneous spherical shell made of currently available materials (please see our US patent application 20070001053 at https://www.google.com/patents/US20070001053 ). The main failure mode is buckling. However, our finite element analysis (please see the same application) showed that it is possible using a spherical (inhomogeneous) sandwich structure, e.g., if the face sheets are made of beryllium and the core is made of commercially available aluminum honeycomb. To avoid the so called intracell buckling, the diameter of the shell made of these materials should be at least 3 m. This requirement can be relaxed using a different core. 
A: There's a nice dimensional argument that says that you can't just consider very large spheres as a solution to this.
Consider an evacuated sphere of radius $R$, and consider cutting it in half.  The force across this cut goes as $R^2$ obviously.  The length of the cut goes only as $R$ though, so in order to keep the stress constant (and below the stress at which the material of the sphere fails), the thickness of the material must go like $R$ as the sphere becomes large.
The total volume of the material the sphere is made of therefore goes like $R^2\times R$, where the first term comes from the area and the second the thickness, as the sphere becomes large.  So the mass of the sphere goes like $R^3$.
But the volume of the sphere, and thus the total lift, also goes like $R^3$.  So, above a certain point if the sphere does not have enough lift then you can't fix this by making it larger.
Note: in practice buckling will be a much more serious problem as another reply said: I'm just putting this here because the existence of a bound is interesting, and it's too long to be a comment.
A: It has been done, but with the help of helium inside the metal to aid in the displacement, when the Mythbusters flew a lead balloon. I know that the question is about an evacuated sphere, and not a balloon, but the difference is, from the point of view of physics, a question of engineering, not science.
A: The weight of the air with a volume equal to the volume of the sphere, where the radius R reaches the inner side of the sphere is:
$\frac4 3\pi R^3\rho_{air}$
The weight of the metal sphere is:
$\frac {4} {3}\pi\rho_{metal} ((R+T)^3-R^3)=\frac 4 3\pi\rho_{metal}(3R^2T+3RT^2+T^3) $
Now
$ P_{crit}= (\frac T R)^2 $, so $T=\sqrt P_{crit}R$  (see the comment above, where T is the thickness of the sphere)
Substituting this in the previous expression gives the weight of the metal sphere:
$\frac4 3\pi\rho_{metal}R^3(3\sqrt(P_{crit})+3P_{crit}+P_{crit}\sqrt(P_{crit}))$
So if we can find a metal (which we haven't (yet)) for which or the density, or the critical pressure, or both are lower than is the case in present metals, we can let the sphere float in the air.
