Balls and Urn Paradox So, I came across the following paradox:

At $1$ minute before noon, put in balls $1 \sim 10$ and take out ball number $1$. At $1/2$ minute before noon, put in balls $11\sim20$ and take out ball number $2$ and so on. How many balls are there at noon?

None.

At $1$ minute before noon, put in balls $1 \sim 10$ and randomly take out a ball. At $1/2$ minute before noon, put in balls $11\sim20$ and randomly take out another ball and so on. How many balls are there at noon?

None.

Okay, so I understand the first paradox because one can describe the exact moment each ball was taken out. But, you can't apply a similar argument to the second paradox because we randomly take out a ball.
  I feel as if it's like infinitely summing $\frac{1}{n}$ and eventually there would be too many balls.

Can someone explain to me mathematically why this is the case?
Let $ n \in \mathbb{N} $

Let the ball labelled $ n $ be placed in the urn on step $ i_0 $.

The probability of removing the ball labelled $ n $ at step $ i $ is $ \dfrac {1} {9i+1} $.

The probability of it never being removed is therefore

$$ \prod_{i=i_0}^{\infty} 1 - \dfrac {1} {9i+1} $$

Which diverges to $ 0 $ since the sum

$$ \sum_{i=i_0}^{\infty} -\dfrac {1} {9i+1} $$ 

diverges to $ -\infty $.

So all balls will be removed at some point with probability $ 1 $.
It depends. How to the laws of the universe work in your hypothetical situation? If the laws of the universe work in such a way that number of balls in the urn is the limit as time approaches noon from the past direction of the number of balls in the urn at that time, then there will be infinitely many balls in the urn at noon, if you're careful what you really mean when you say that as I will describe later. If the laws of the universe work in such a way that each individual ball will be in the urn at noon if and only if it was in the urn at all times before noon that are sufficiently close to noon, then there will be none in the urn at noon.

Now I'll explain what I mean when I say the number of balls in the urn at noon is the limit as time approaches noon from the past direction of the number of balls in the urn at that time. In Calculus, there is only one kind of infinity. In set theory, there are multiple infinities. In set theory, a cardinal number is used to describe an equivalence class of sets that all of a bijection from one another, and it can be shown that there are multiple infinite cardinal numbers. When the limit as time approaches noon from the past side of number of balls in the urn at that time is the Calculus infinity, I declare it correct to say the number of balls in the urn at noon is the limit when the number of balls in the urn at noon is $\aleph_0$, the number of natural numbers there are, in the context of this answer.

Now I'll explain how it's possible that if each individual ball will be in the urn at noon if and only if it was in the urn at all times before noon that are sufficiently close to noon, there will be none in the urn at noon. Suppose each individual ball is in the urn at noon if and only if it was in the urn at all times before noon that are sufficiently close to noon. Consider the time when so far, you just put in balls 1-10 and took out ball one. At that time, there are just balls 2-10 in the urn. Eventually all the balls from 2-10 will be taken out when you put in balls 90-100 and take out ball 10. The balls currently in the urn at this time, 11-100 also will eventually be all taken out when you put in balls 990 to 1,000 and take out ball 100.

Now we can conclude that each individual ball some time before noon will be taken out and never put back in any time before noon. Given what the laws are, that means none of the balls will be in the urn at noon. However, as time approaches noon from the past direction, the number of balls in the urn approaches infinity. This shows that the number of balls in the urn at noon is not the limit as time approaches noon from the past direct of the number of balls in the urn at that time.