Blowing resistors in parallel, in an order-controlled way
If the resistors are all the same package and wattage, they should blow in order of high to low abuse.  In this case abuse would be dumping too much power thru them.  The power dissipated by a resistor is V**2 / R.  Since the resistors are in parallel and V therefore the same for all, those with smaller R will suffer proportionately higher abuse.

So, arrange them in order from low to high resistance.  The existance of Rs will cause the voltage accross the resistors to go up each time one pops, hastening the demise of the next one in line.  This also means you should calculate each value so that it dissipates the necessary power to pop with all previous resistors open.  Note that Rs needs to be quite beefy so as to not itself pop.

Let's say you have determined that 1 W dissipation will cause the desirable popping in the types of resistors you plan to use and that Vs is 12 V (a car battery would work well as it is a good voltage and can easily handle the power).  Let's also say that when only the last resistor is left, Rs drops 1 V.

To calculate the cannon-fodder resistors, work backwards from the last.  When only the last resistor is left, it will have 11 V applied to it.  Since we want 1 W dissipation, the resistance in Ohms will be the square of the Volts applied to it, which is 121 &Omega; for the last one.  This also tells you that Rs must be 11 &Omega;.

Now you can calculate the value for the second to last resistor.  The Thevenin equivalent it sees is 10.08 &Omega; and 11 V.  So the question is what resistance connected to that Thevenin source dissipates 1 W?  The equation is a quadratic, which I'll leave for you to solve.  Once you have that resistance, you can calculate the Thevenin source the next resistor sees and repeat the process as far as you like.