Life In 19x19

Willemien ( killing lines ) inspired me to raise another problem.
You play even against your friend at n handicaps ( she playing black ).
Someday she proposes that instead of handicaps you play without ko and zero komi. In the next game black is always allowed to immediately retake in a ko but white isn't (*).
What is the lowest n for which you would consider the proposal as profitable for white?

(* To avoid discussion maybe it is better to state that white is not allowed to recreate a previous position but black is.)

I don't like that formulation where you equate handicap stones to no-ko since this has the underlying assumption of different playing strengths of players, i.e. it relies on suboptimal play of black.
I'd reformulate it as:
What is the correct komi if white is not allowed to repeat previous positions but black is allowed to? This way you can use equally strong players, and in theory even perfect play.

Li Kao wrote:I don't like that formulation where you equate handicap stones to no-ko since this has the underlying assumption of different playing strengths of players, i.e. it relies on suboptimal play of black.
I'd reformulate it as:
What is the correct komi if white is not allowed to repeat previous positions but black is allowed to? This way you can use equally strong players, and in theory even perfect play.

I can't disagree. But it seems too late to reformulate so I drop the topic. The replies this far indicate a few handicap stones so maybe 30 or 40 komi.