Your IQR will cover the middle 33 observations. Set all of those equal to zero. Then the IQR is zero, and any nonzero point will be an outlier. Thus, half of the points can be "outliers" by your definition.
(There is some funkiness because of the even number of observations, but there are many different ways to calculate empirical IQR, and this is the gist. Because of the even number of observations, it seems that the answer is 32 points instead of 33, and that the general solution is going to be:
$$\left \lfloor{\dfrac{n - 1}{2}}\right \rfloor .$$
(That is, round down the $x.5$ number for an even $n$, or take the integer value for an odd $n$.)
x <- c(rep(-100, 16), rep(0, 34), rep(100, 16)
y <- c(rep(-100, 17), rep(0, 33), rep(100, 17)
boxplot(x, y)