24. (IND 1) An odd integer $n \geq 3$ is said to be "nice" if there is at least one permutation $a_{1}, a_{2}, \ldots, a_{n}$ of $1,2, \ldots, n$ such that the $n$ sums $a_{1}-a_{2}+$ $a_{3}-\cdots-a_{n-1}+a_{n}, a_{2}-a_{3}+a_{4}-\cdots-a_{n}+a_{1}, a_{3}-a_{4}+a_{5}-\cdots-a_{1}+$ $a_{2}, \ldots, a_{m}-a_{1}+a_{2}-\cdots-a_{n-2}+a_{n-1}$ are all positive. Determine the set of all "nice" integers.
24. Let $y_{k}=a_{k}-a_{k+1}+a_{k+2}-\cdots+a_{k+n-1}$ for $k=1,2, \ldots, n$, where we define $x_{i+n}=x_{i}$ for $1 \leq i \leq n$. We then have $y_{1}+y_{2}=2 a_{1}, y_{2}+y_{3}=$ $2 a_{2}, \ldots, \psi_{n}+y_{1}=2 a_{n}$.
(i) Let $n=4 k-1$ for some integer $k>0$. Then for each $i=1,2, \ldots, n$ we have that $y_{i}=\left(a_{i}+a_{i+1}+\cdots+a_{i-1}\right)-2\left(a_{i+1}+a_{i+3}+\cdots+a_{i-2}\right)=1+$ $2+\cdots+(4 k-1)-2\left(a_{i+1}+a_{i+3}+\cdots+a_{i-2}\right)$ is even. Suppose now that $a_{1}, \ldots, a_{n}$ is a good permutation. Then each $y_{i}$ is positive and even, so $w_{i} \geq 2$. But for some $t \in\{1, \ldots, n\}$ we must have $a_{t}=1$, and thus $y_{t}+y_{t+1}=2 a_{t}=2$ which is impossible. Hence the numbers $n=4 k-1$ are not good.
(ii) Let $n=4 k+1$ for some integer $k>0$. Then $2,4, \ldots, 4 k, 4 k+1,4 k-$ $1, \ldots, 3,1$ is a permutation with the desired property. Indeed, in this case $y_{1}=y_{4 k+1}=1, y_{2}=y_{4 k}=3, \ldots, y_{2 k}=y_{2 k+2}=4 k-1$, $\zeta_{2 k+1}=4 k+1$.

Therefore all nice numbers are given by $4 k+1, k \in \mathbb{N}$.