Six. (20 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$. Given
$$
\begin{array}{l}
a_{k}=\sum_{i=k^{2}}^{(k+1)^{2}-1} \frac{1}{i}(k=1,2, \cdots) . \\
\text { Find } \sum_{k=1}^{n}\left(\left[\frac{1}{a_{k}}\right]+\left[\frac{1}{a_{k}}+\frac{1}{2}\right]\right) .
\end{array}
$$
Six, first prove: for any $x \in \mathbf{R}$, we have
$$
[x]+\left[x+\frac{1}{2}\right]=[2 x] \text {. }
$$

In fact, when $0 \leqslant x-[x]<\frac{1}{2}$,
$$
\left[x+\frac{1}{2}\right]=[x],[2 x]=2[x] \text {. }
$$

Then $[x]+\left[x+\frac{1}{2}\right]=2[x]=[2 x]$.
When $\frac{1}{2} \leqslant x-[x]<1$,
$$
\left[x+\frac{1}{2}\right]=[x]+1,[2 x]=2[x]+1 \text {. }
$$

Then $[x]+\left[x+\frac{1}{2}\right]=2[x]+1=[2 x]$.
Next, prove: for any $k \in \mathbf{Z}_{+}$, we have
$$
\frac{2}{k+1}<a_{k}<\frac{2}{k} \text {. }
$$

In fact, the expression for $a_{k}$ has $2 k+1$ terms. Dividing it into two parts, the sum of the first $k$ terms and the sum of the next $k+1$ terms, we denote
$$
a_{k}=a_{k_{1}}+a_{k_{2}},
$$

where $a_{k_{1}}=\sum_{i=k^{2}}^{k^{2}+k-1} \frac{1}{i}, a_{k_{2}}=\sum_{i=k^{2}+k}^{(k+1)^{2}-1} \frac{1}{i}$.
$$
\begin{array}{l}
\text { By } \frac{1}{k+1}=k \cdot \frac{1}{k^{2}+k}<a_{k_{1}} \leqslant k \cdot \frac{1}{k^{2}}=\frac{1}{k}, \\
\frac{1}{k+1}=(k+1) \frac{1}{(k+1)^{2}}<a_{k_{2}} \\
<(k+1) \frac{1}{k^{2}+k}=\frac{1}{k},
\end{array}
$$

we know $\frac{2}{k+1}<a_{k}=a_{k_{1}}+a_{k_{2}}<\frac{2}{k}$.
From equation (2), we get $k<\frac{2}{a_{k}}<k+1$. Then $\left[\frac{2}{a_{k}}\right]=k$.
From equation (1), we get
$$
\begin{array}{l}
{\left[\frac{1}{a_{k}}\right]+\left[\frac{1}{a_{k}}+\frac{1}{2}\right]=\left[\frac{2}{a_{k}}\right]=k .} \\
\text { Hence } \sum_{k=1}^{n}\left(\left[\frac{1}{a_{k}}\right]+\left[\frac{1}{a_{k}}+\frac{1}{2}\right]\right) \\
=\sum_{k=1}^{n} k=\frac{n(n+1)}{2} .
\end{array}
$$