You would need 100 different jars with different counts.
Otherwise, you either get lucky on the first time and then you always know the amount, or you (more likely) loose the first time and then it doesn't matter.
No, you don't need 100 different jars. Let's say the jar started with S beans, and the person grabbed X_1, X_2, .. X_n beans in n trials. To prove the superpower, you have to tell him the correct value of X_i every time. Knowing S beforehand does not help you unless you can calculate S-X-i without knowing X_i.
I think you're assuming that you can count the jar between pulls. (That was my initial reading also.) In that case, you guess your first number, count the jar, and then work backwards for all subsequent pulls.
But on rereading, the info to prove isn't "I know how many beans were here to start", it's "I'm capable of counting this jar of beans by looking at it". So if you guess right on the first trial, you still won't know the beans-remaining count, and will have to guess on the next trial too.
This is made super confusing by the claim that it's a ZKP of the initial count. That's narrowly true, but the real question at hand is whether you can count the beans.
Otherwise, you either get lucky on the first time and then you always know the amount, or you (more likely) loose the first time and then it doesn't matter.