If you have a source of radial magnetic field $B\sim Q_M/r^2$, then one may prove that the vector potential $\vec A$ can't be single-valued. It's because $\vec B={\rm curl}\vec A$ for a well-defined $\vec A$ automatically satisfies ${\rm div}~\vec B=0$. However, $Q_M/r^2$ has a curl proportional to the delta-function at the origin.
Still, this delta-function vanishes everywhere except for a Dirac string (and the space minus the Dirac semiinfinite string is simply connected), so with the Dirac string, $\vec A$ may be defined everywhere. $\vec A$ still changes under the loop around the Dirac string. In this way, the magnetic monopole is replaced by a very long magnetic dipole. Two poles are connected with a thin solenoid and one of the monopoles is sent to infinity and becomes irrelevant. The Dirac string, i.e. a very thin solenoid, becomes unobservable as well, even for interference experiments (as long as the confined magnetic flux is properly quantized).
The arguments above are waterproof and you can't circumvent them. So if you're asking whether there is a way to introduce a magnetic monopole so that the vector potential would be single-valued, the answer is a resounding No, much like if you ask if it is possible to introduce the number 4 so that it isn't equal to 2+2.
However, one may try to search for solutions to similar, less singular problems. In theories with Higgses, one may "dilute" the delta-function a little bit and find non-singular solutions of Yang-Mills theories with Higgs fields, the so-called
http://en.wikipedia.org/wiki/%27t_Hooft%E2%80%93Polyakov_monopole
't Hooft-Polyakov monopole that is non-singular but indistinguishable from the Dirac monopole when you're very far from the center of the solution, relatively to its characteristic length scale. This solution has various generalizations, too.