How much should I pay for a chance to win 100$?

You can find the money behind door 1,2,3 or 4 with equal probability and if the price is $X$ to open a door, your winnings are $(100-X)$, $(100-2X)$, $(100-3X)$ or $(100-4X)$ depending what door you find the money behind.

Therefore your expected return is $$\frac14\bigl((100-X)+(100-2X)+(100-3X)+(100-4X)\bigr) = 100 -2.5X$$

For you to expect to make a profit then $X$ must be less than $\$40$. So you would make a profit over many games if $X$ is less than $\$40$.

If you decide to stop when you find the $\$100$ or on the $n$th door, then your return is $$\sum_{i=1}^n\frac14(100-iX)+\left(1-\frac n4\right)(-nX)=\frac n8(200+X(n-9)).$$ Let this be greater than $0$ to find that $$X<\frac{200}{9-n}$$ which is maximised when $n=4$ so $X<\$40$.