You have 50 bags. How many marbles do you need at least, so that you can have a different number of marbles in each bag?
The minimum number of marbles is:
$49$. Put a single marble in $49$ of the bags, and then put all the bags one inside each other, and then put the empty bag inside them all. In other words, going from outside to in, we will have bag-marble-bag-marble-...-bag-marble-bag, and the sequence of the number of marbles in each bag going from inside to out will be $0,1,2,3,...,49$.
We can prove this is the minimum number because:
For there to be a different integer number in each bag, the sequence that gives the smallest possible number in the bag with the most marbles is $0,1,2,...,49$, so at least one bag must have $49$ marbles in it.
If we increase the laterality of our thinking even further (with weakened respect for dimensional constraints, and tongue lightly pressed in cheek), we can improve our solution to:
$48$ marbles. Set up the first $49$ bags as above with the $48$ marbles, so that they contain $0,1,2,...,48$ marbles each. Then alter the final bag so that it has the topology of a Klein bottle, and set it down where you please; it now contains every marble in the universe, since its interior is indistinguishable from its exterior. Our bags thus contain $0,1,2,...,47,48,|M|$ marbles where $M$ is the set of all objects in the universe that could properly be construed as marbles, whose size (given that I also happen to own a marble, and you have $48$) is strictly greater than $48$.
Additionally, with tongue now firmly planted in cheek, the minimal number can be drastically reduced to:
$1$ with a simple repeated application of the Banach-Tarski theorem.