Expected number of samples to get a number greater than X?

Consider a population of people. Each person's height is IID (you're not given the distribution). You randomly choose a person and observe that their height is $$X$$. Let $$N$$ be the number of additional samples needed to randomly select a person who's height is greater than $$X$$. What is $$E[N]$$?

Apparently, $$E[N] = \infty$$ and the logic is that $$X$$ is a random variable.

My first concern is that I am not sure why $$X$$ is a random variable. Once you obtain a sample with a value $$X$$, isn't $$X$$ now fixed/observed (the instructions state the height is observed) and no longer random? In that case, the expectation should be bounded?

If we assume $$X$$ is indeed a random number. Then I believe you do the following

$$E[N] = \sum_i E[N|X = X_i] P(X_i) =$$

We apply the law of total expectation and condition on all possible values of $$X$$. Is the expectation infinity because $$X$$ can take on the value $$\max(heights)$$, in which case you will never get a person with a height greater than $$> \max(heights)$$ no matter how long you sample for? If there was a constraint that you know $$X$$ isn't $$\max(heights)$$, then in this case, would $$E[N]$$ be bounded?

Suppose you select the tallest person in the population; there is an assumption that the population is finite. There is a nonzero probability that this happens. If you select the tallest person, you'll never find a higher one, so $$N=\infty$$. Since $$\infty×p=\infty$$ if $$p>0$$, when $$X$$ is treated as a random variable on which $$N$$ depends, we have $$E[N]=\infty$$.
Let's say you need $$n$$ people to find the one who is taller than the first. You sort these $$n+1$$ people by height and it happens that the first one got into the place number $$n$$ and the last one got into the place number $$n+1$$. The probability of this is $$P\{N=n\}=\frac{(n-1)!}{(n+1)!}=\frac1{n(n+1)}$$ Therefore the expectation is indeed infinity $$E[N]=\frac12+\frac13+\frac14+...=\infty$$