Here is the general approach you should take.
Any connectors and contact points to the outside world should have protection. The outer perimeter of the board should have a guard ring running around it with protection on it. This is because people tend to handle the board by the edges. This guard ring should be the ground for the rest of the system with a another ring concentric just inside.
All other internal nodes probably can't be protected, in fact bare boards are never considered to be ESD proof for all possible scenarios.
You can encapsulate the board to provide a little more protection, depending upon the encapsulation it can be quite good. 32 um of parylene would really help but is expensive (this is a mil standard).
But ultimately, the whole board should be inside of an enclosure to prevent contact and discharge.
The first thing that comes to mind is that ESD testing should be done on points that are normally accessible during normal operation. Since you haven't described what you are building, in your question it makes up for kind of a vague answer. The testing is also not described in your question, and the most important thing is what points the ESD gun is connected to. The human body model is with respect to earth ground. 
When you say positive strike, do you mean positive with respect to earth ground? Positive with respect to what? The current is going to take the lowest impedance path back to the source. You have to connect the negative or ground side of the ESD gun to something. Normally it is connected to earth ground or chassis ground. The current must travel back to the source, so if you hit the board with the ESD gun connected to ground, it has several 'potential' paths that it could take back. There is the usb shield, the usb ground, the D+ and D- wires, Vbus and the air. The Air is like a 10^6 to 10^9 resistor with a few pF's of capacitance. The D+ and D- wires have at least the resistance of Rt plus whatever is on the driving end of the cable. Vbus would probably have a voltage regulator on it so it would be less than that of Rt. The ground has the resistance of the cable and the connectors, which is probably the lowest of all, being an ohm or two to earth ground (since ground and earth ground are connected somewhere down the line).
So the answer is, its going to take the earth ground back to the source (if you have the esd gun connected to ground, and all paths will see it to some extent, the ground will take most of the current)
Back to the design:
The whole reason for having a shield around USB is to shunt noise and ESD away from the grounding wire that is next to the V+ and V- cables. Because of mutual inductive coupling, you could get some transients that cross over to the datalines and at a minimum create noise that knocks out the packet that was sent.
The best thing would be to use the shield to shunt noise and ESD away from the datalines and usb ground. The next best thing (and not recommended) would be to shunt the ESD down the ground wire of the usb cable.
If the pcb is being handled by humans try and design the pcb so ground is the first thing seen by an ESD event. The ideal situation would be to have a metal case that encloses your device tied to the usb shield. If you can't tie the enclosure to the shield (not recommended) tie it to ground. If you cant have an enclosure then try and use guard rings and a nice healthy ground plane.
Best Answer
Have you read this article from Texas Instruments ?
To test your ESD protection properly you need to create the correct pulse. This can be done using a circuit like this:
As you can see the idea behind the Human Body ESD model is that a capacitor is charged to a certain voltage and that charged capacitor is then discharged (through a defined series resistor) through the device under test.
The discharge of the capacitor causes a current to flow and a voltage to develop across your DUT. You need to check if that current and voltage do not exceed the limits of what your DUT (the protection inside it) can handle. At least, that's how I simulate my ESD protections!
Your circuit will only briefly charge the capacitors, it does not simulate the high voltage discharge at all.