10. (20 points) Given $n(n>1)$ is a positive integer, let $a_{k}=\left(\frac{n-k}{k}\right)^{k-\frac{n}{2}}(0<k \leqslant n, k \in \mathbf{N})$. Find the maximum value of the sequence $\left\{a_{k}\right\}$.
10. Let $f(k)=a_{k}(0 \leqslant k \leqslant n, k \in \mathbf{N})$.
$$
\begin{array}{l}
\text { Given } \ln f(k)=\ln \left(\frac{n-k}{k}\right)^{k-\frac{n}{2}} \\
=\left(k-\frac{n}{2}\right) \ln \frac{n-k}{k}=\frac{1}{2}(2 k-n) \ln \frac{n-k}{k} \\
=-\frac{1}{2}(k-(n-k))(\ln k-\ln (n-k)) .
\end{array}
$$

Notice that, $k-(n-k)$ and $\ln k-\ln (n-k)$ have the same sign.

When $k>n-k$, i.e., $k>\frac{n}{2}$, $k-(n-k)$ and $\ln k-\ln (n-k)$ are both monotonically increasing and always positive with respect to $k$, thus, the closer $k$ is to $\frac{n}{2}$, the smaller the value;

When $k<n-k$, i.e., $k<\frac{n}{2}$, $k-(n-k)$ and $\ln k-\ln (n-k)$ are both monotonically increasing and always negative with respect to $k$, thus, the closer $k$ is to $\frac{n}{2}$, the smaller the value.

This indicates that when $k$ takes the integer closest to $\frac{n}{2}$, $(k-(n-k))(\ln k-\ln (n-k))$ reaches its minimum value, i.e., $\ln f(k)$ reaches its maximum value, i.e., $f(k)$ reaches its maximum value.
Therefore, when $n \equiv 1(\bmod 2)$,
$$
f(k)_{\max }=\max \left\{f\left(\frac{n-1}{2}\right), f\left(\frac{n+1}{2}\right)\right\}=\sqrt{\frac{n-1}{n+1}} \text {; }
$$

When $n \equiv 0(\bmod 2)$,
$$
f(k)_{\max }=f\left(\frac{n}{2}\right)=1 \text {. }
$$

Thus, when $n \equiv 1(\bmod 2)$, the maximum value of $\left(\frac{n-k}{k}\right)^{k-\frac{n}{2}}$ is $\sqrt{\frac{n-1}{n+1}}$; when $n \equiv 0(\bmod 2)$, the maximum value of $\left(\frac{n-k}{k}\right)^{k-\frac{n}{2}}$ is 1.