Resistive braking of series wound DC motor
The torque of an ideal series wound DC machine is \$T = k i^2\$.  So the torque is always in the same direction, regardless of the current direction.  

Regardless of how you're generating current, if you want to reverse the direction of the torque you need to change the constant \$k\$.  You can only do that by changing the polarity of the field coil with respect to the armature.

So, if you want to brake -- by any means -- you need to reverse the direction of the field coil.

** Edit **

Going to the next level of detail, by conservation of energy*, in an ideal motor, \$v_a i = T \omega\$, where \$v_a\$ is the armature voltage and \$\omega\$ is the shaft speed.  So \$v_a = k i \omega\$.

Model the motor with some inductance, so the inductive contribution is \$v_L = L \frac{di}{dt}\$.

\$E = I\, R\$, and if you're careful with your sign conventions you'll see that the resistor opposes the coil voltage, so \$i = -\frac{v_a + v_L}{R}\$.

Separating that out, and doing more work in one step than would be acceptable in a homework assignment: $$i = -\frac{k \omega}{R} i - \frac{L}{R} \frac{di}{dt}$$

Rearrange terms and again doing more steps than I ought to, I get $$\frac{di}{dt} = - \frac{R + k \omega}{L} i$$.

So -- if you just put a resistor on the thing, the current will decay exponentially based on the speed of the motor; i.e., the ideal motor running at constant speed "looks" like a resistor.  This makes sense for energy conservation, because \$P_{out} = T \omega\$, and \$P_{in} = i^2 R\$, and the energy stored in the inductor is a function of \$i^2\$.  Since the power out is a function of the shaft speed, it would make sense that the energy in the inductance would get sucked out faster the faster the motor is turning.

I didn't try modeling the dynamics of the motor itself (\$\dot \omega = T/J\$, with the angular acceleration, \$\dot \omega\$, depending ultimately on current) because I assume that they are slow enough that the motor current decays long before the motor speed has a chance to.  So for the purpose of this problem, you can take the speed as constant.

* Thank you Emily Noerther!  This makes it so easy!