Prove that the conclusion is obviously true when $n=1$. Let $n \geqslant 2$, and denote the said sum as $A$, then
$$2 A=2+\left(2^{k}+n^{k}\right)+\left(3^{k}+(n-1)^{k}\right)+\cdots+\left(n^{k}+2^{k}\right) .$$

Since $k$ is a positive odd number, by the factorization formula (6), for each $i \geqslant 2$, the number $i^{k}+(n+2-i)^{k}$ is divisible by $i+(n+2-i)=n+2$, hence the remainder when $2 A$ is divided by $n+2$ is 2, thus $A$ cannot be divisible by $n+2$ (note that $n+2>2$).