That circuit would be very marginal even with red LEDs. Blue LEDs have a higher forward voltage (Vf) than red ones, so you would need to arrange them as 2 parallel strings of 2, rather than 1 string of 4 as shown. Each string would need its own current-limiting resistor. The resistor value would be given by (9-2*Vf)/I, where Vf is the blue LED forward voltage and I is the required LED current in Amps.