TL; DR: the solution is
Tim starts with $0$ marbles, Ava starts with $17$ marbles, and Jack starts with $44$ marbles.
Let $t$, $a$, and $j$ be the initial number of marbles that Tim, Ava, and Jack have, respectively. Let's go through the story and write it in mathematical language.
Tim: "Wow, marbles! Lets share!"
Ava: "Yeah! We found them together, we should share equally. One for your pile, one for my pile, another for your pile, another... There! We're equal!"
Now, Tim has $t+j/2$, Ava has $a+j/2$, and Jack has $0$.
This gives us the first condition:
As no marbles are left, $j$ is even
Tim: "No, I saw them first, I'll have two more than you! I'll take this, and this!"
There is a problem here. If Tim takes two marbles from Ava, then Tim won't have two "new" marbles more than Ava, but 4. I assume that Tim does not know this, and that he indeed takes two marbles from Ava's pile.
Now, Tim has $t+j/2+2$, Ava has $t+j/2-2$, and Jack has $0$
This gives us the second condition:
Ava: "That's not fair!"
Jack: "Hey! What are you two doing with my marbles?"
Tim: "Your marbles?"
Jack: "Yeah, I put them there!"
Tim: "Sorry, I already dumped my fair share of marbles into my marble sack, and now I don't know how many are yours."
Ava: "Yeah, sorry, I dumped my unfair share of marbles into my marble sack too."
Jack: "Awww... Wait, I have an idea! You and Tim can simply give me half of the number of marbles in your bags, that should be fair."
Tim: "One for your pile, one for my pile, another for your pile, another... There!"
Now, Tim has $t/2+j/4+1$, Ava has $a+j/2-2$, and Jack has $t/2+j/4+1$.
This gives us our third condition:
No leftover marbles, so $t+j/2+2$ is even, or, equivalently, $t+j/2$ is even.
Ava: "One for your pile, one for my pile, another for your pile, another... Here's one left over!"
Tim: "I'll take that. In my pile it goes!"
Now, Tim has $t/2+j/4+2$, Ava has $(a-1)/2+j/4-1$, and Jack has $t/2+(a-1)/2+j/2$.
This gives us our fourth condition:
$a+j/2-2$ is odd, or, equivalently, $a+j/2$ is odd.
Ava: "Hey! Why your pile?"
Tim: "Because it's an extra, right? Nobody's losing anything."
Jack: "Ummm... this is way less than I remembered to have!"
This gives us our fifth condition:
$t/2+(a-1)/2+j/2\ll j$. Equivalently, $t+a-1\ll j$.
Ava: "Sorry, that's the best we can do for you, right Tim?"
Tim: "Yeah, take it or leave it."
Jack: "This is crazy! Return my marbles!"
Ava: "Hey! Get your hands off my marbles!"
Tim: "Hey! Mine too!"
Jack: "Too late, I dumped the marbles into my marble sack, bye!"
Now, they all have the same number of marbles as they had at the beginning. This means that Jack took from Tim
and Jack took from Ava
Jack took from Tim 12 more than from Ava, which gives us our sixth condition:
and our seventh condition:
Jack must take at least one marble from Ava, so
Now it is time to find the minimal $t$, $a$, and $j$ that satisfy all seven conditions:
- None of the conditions has $t$ on the large side of an inequality, so we set $t=0$ and see where that takes us.
- From the sixth condition, we have $a=17$.
- From the seventh condition, we have $j\ge 44$. We set $j=40$ and see what happens.
- First condition: $j$ is even: check.
- Second condition: $t+j/2\ge 2$. We have $t+j/2=22$, so check.
- Third condition: $t+j/2$ is even: check.
- Fourth condition: $a+j/2$ is odd. We have $a+j/2=39$, so check.
- Fifth condition: $t+a-1\ll j$ We have $16\ll 44$, so check.
All conditions are satisfied, so the solution is
$t=0$, $a=17$, $j=44$.
If you follow the story again, everything adds up. In the last step,
Jack takes $13$ from Tim, and Jack takes $1$ from Ava, going from (Tim:13, Ava:18, Jack:30) to the final stage (Tim:0, Ava:17, Jack:44), which is equal to the initial stage.