Assuming that the 27 watts is important to the function of the relay, I believe that the installation of a resistor in parallel with the blinkers would work. Someone else may be able to tell you whether or not the power drawn by the LED's is a factor, but I think the following would increase the wattage drawn through the relay, hopefully making it work.

Since the new lights are 12v and have two different "filaments", one that's 8w and the other that is 23w you will need a different resistor depending on what side you are using. Note that Watts = Volts * Amps

Use this formula to determine the resistance required:

voltage^2 / (watts requried - bulb watts) = resistor

So if you used the 23w side, you would need (12v^2) / (27w-23w) ~ 36 ohms of resistance.

If you use the 8w side, 144/(27-8)~8 ohms of resistance.

This formula works because loads in parallel (the LED and the resistor) will have the same voltage but will vary in current. As a result, the bulb will get it's requisite voltage and current, while the resistor will pull additional current and therefore increase the watts "seen" by the relay.

(1) V = I*R

(2) W = V*I

(W is watts V is volts, I is current (amps) and R is resistance)

If we use equation (2) and solve for I, we get W/V = I. Next we can plug I = W/V into equation (1). This gives us V = W/V * R. Solving for R we get V^2/W = R. We want the relay to "see" 27 watts, we need the relay to pull the rest of the watts that the LED wont be pulling so that's why we do (watts requried- bulb watts) and due to the LED and the resistor being in parallel, they both have 12 volts.

I'm not sure how much you know about circuits but when something is installed in parallel, it will be connected across the LED. This means that one side of the resistor would be connected to the ground of the LED and the other would be connected to the "hot" wire coming into the LED. In series, the resistor would be put in, replacing a section of the positive lead coming into the LED.

Sorry for the long post, I hope it helps.

TL;DR

Installing a resistor with resistance R in parallel with the LED should increase the power drawn from the relay.

Where:R = 144 / (27-LED wattage)