Sharing the load between resistors, parallel or series?
In order to "share the load" between resistors, each resistor must be able to dissipate half the desired wattage. 

In the case of the question, if the desired resistance is R and wattage W, then putting two (ideal) resistors of value 2 x R each in parallel will result in a resistance of R, but with each of the resistors passing only half of the current.



For a parallel combination of resistors, the resultant resistance is given by:
R = 1 / ( 1/R1 + 1/R2)

Thus, to achieve a value of R using a parallel combination of two resistors, we must use R1 = R2  = 2 x R

Now to calculate the power:
Power P = V^2 / R  

The the voltage across the resistor combination is constant, V
Power through resistor R1 from above: = P1 = V^2 / (2 x R) = P / 2
Similarly P2 = V^2 / (2 x R) = P / 2

Thus, each resistor will dissipate half the specified power, or in other words will need to be half the stated power rating W.



In reality, resistors differ in actual resistance due to manufacturing tolerances. Hence, one resistor might carry slightly more current than another, both being nominally of equal value. To provide headroom for such variations, it is best to allow for slightly higher power ratings for the resistors, than is absolutely required.