Hmm, I'm sure I've seen this problem before but I remember I can't solve it at that time

see Dinitz's Theorem

Where can I find it ?

I solved it in the exam, it took about 2 hours fully. No, I never heard of this problem before. I first noticed that basically it should be somehow traced back to Hall's Marriage Theorem (which I worked with for half a month after a math camp and managed to find a fully own proof, I randomly noticed that a statement like this should be true if another exercise was true so I looked it up and found this theorem), the question was how to manage to pair up the two problems. The solution was roughly this: we try to go row by row and try to permute the numbers in a proper way, we'll show that for each row it is possible, no matter how we arranged the numbers above. Permute the first row's numbers in a random way. Going from up to down in the rows, assume the $k$-th row can't be properly filled, we have a look at it. Draw a a bipartite graph such that one part is the numbers in the $k$-th row and the other part is the cells in the $k$-th row. We connect the numbers with the cells for which they can be placed in (which depends on the $k-1$ rows above, if for example one of the rows contained a number in the $t$-th column above, then we can't place that in the cell in the $t$-th column and $k$-th row). Now, we know Hall's Marriage Theorem so we assume that there are some $m$ amount of numbers that are only connected to $l$ amount of cells and that $m>l$. We note and proof the following statements: -a number is connected to at least $n-(k-1)$ cells -a cell is connected to at least $n-(k-1)$ numbers -if a cell is connected to more than $n-(k-1)$ numbers then it is connected to a number which didn't appear in all the rows above (it's possible that it appeared in some, but not all) With these, to show that our statement that $m>l$ is false, we do the following moves: -Search for cells that have more than $n-(k-1)$ edges. By the statement that it is connected to at least one number that didn't appear in all the rows above, we're basically saying that it is connected to a number that also has more than $n-(k-1)$ edges. Delete that connection. If the cell still has more than $n-(k-1)$ edges then it's still connected to a number that has more than $n-(k-1)$ edges, delete that too. Keep doing till we get that the cell has exactly $n-(k-1)$ connections. One thing to notice, we know that each cell is connected to at least $n-(k-1)$ numbers so after this algorithm, we will have exactly $n-(k-1)$ edges for all cells. Other thing to notice, when we take away an edge, we won't make any number have less than $n-(k-1)$ edges. That is because we always take away 1 edge from a number that has more than $n-(k-1)$ edges. The last thing to mention is that because at the start we had at least $n-(k-1)$ edges at all numbers, we'll still have atleast $n-(k-1)$ edges at all numbers. Now, counting all the edges, (denote it $S$), we first get that $S=l*(n-(k-1))$, and second off that $S\geq m*(n-(k-1))$, from which $l*(n-(k-1))\geq m*(n-(k-1))$, and from this we get that $l\geq m$, which is contradiction. (If you look closely, we can leave the Reductio ab absurdum and just directly prove the same way that for any $m$ amount of numbers, there are at least $m$ amount of cells.) But because of this, we can use Hall's Marriage Theorem to make sure we'll get a proper solution for the $k$-th row. Keep going till you get to the last row then you're done. (Look up the theorem on YT, although my proof is just simple contradiction with an easy algorithm. I also wrote my proof on the theorem on he exam paper for like 30 minutes.) I also had other ideas on proving it with tracing it back to a special Latin square, but it had quite a few holes in it (with unproven known theorems, that I knew were true but didn't know the proof to it, and I think there was a not totally true statement I wrote too). I don't think I'll get much on that, but that's a shame because I spent a good time on the second solution too instead of solving the 2nd problem. I also messed up in the first problem, I misunderstood the question, I still got that the centroid will be the solution but instead of $p(Z)=1/4$ I got $p(Z)=1$.. If I didn't mess that up then I'd prob. race for 1st or 2nd place, nevermind now

Unfortunately, a significant detail in the above solution is wrong, i.e. the statement: misinnyo wrote: -if a cell is connected to more than $n-(k-1)$ numbers then it is connected to a number which didn't appear in all the rows above (it's possible that it appeared in some, but not all) is not true, and this failed the whole argument after that. At least for me, it seem to can't be fixed. For the proof of Dinitz conjecture, which was first proved by Fred Galvin in 1994, I suggest reading this paper which clearly explain it.

Official Solution (In Hungarian)