12. A competition has $n \geqslant 2$ participants, and each day the scores of the participants are exactly the set $\{1,2, \cdots, n\}$. If at the end of the $k$-th day, the total score of every pair of participants is 52 points, find all pairs $(n, k)$ that make this possible.
12. On the one hand, the total score of all players at the end of $k$ days is $k(1+2+\cdots+n)=\frac{1}{2} k n(n+1)$. On the other hand, since the total score of any two players at the end of $k$ days is 52 points, the total score of all players at the end of $k$ days is also equal to $\frac{1}{n-1} C_{n}^{2} \cdot 52=26 n$, thus $\frac{1}{2} k n (n+1)=26 n$, which simplifies to $k(n+1)=52$, leading to the pairs $(n, k)=(51,1),(25,2),(12,4),(3,13)$. It is not difficult to verify that among these, only $(51,1)$ is not a solution to the problem, hence the required pairs $(n, k)$ are only 3: $(25,2),(12,4),(3,13)$.