$34 \cdot 1$ For integers $k$ and $n$, where $1 \leqslant k<n$, let $C_{n}^{k}=\frac{n!}{k!(n-k)!}$, then $\left(\frac{n-2 k-1}{k+1}\right) C_{n}^{k}$ is an integer under the condition:
(A) For all $k$ and $n$.
(B) For all even values of $k$ and $n$, but not for all $k$ and $n$.
(C) For all odd values of $k$ and $n$, but not for all $k$ and $n$.
(D) If $k=1$ or $n-1$, but not for all odd values of $k$ and $n$.
(E) If $n$ is divisible by $k$, but not for all even values of $k$ and $n$.
(27th American High School Mathematics Examination, 1976)
［Solution］Since $C_{n}^{k}$ is an integer, consider
$$
\begin{aligned}
\left(\frac{n-2 k-1}{k+1}\right) C_{n}^{k} & =\frac{n!(n-k)}{k!(n-k)!(k+1)}-C_{n}^{k} \\
& =\frac{n!}{(k+1)!(n-k-1)!}-C_{n}^{k} \\
& =C_{n}^{k+1}-C_{n}^{k} .
\end{aligned}
$$

It is also an integer.
Therefore, the answer is (A).