For any $a, k \in \mathbf{N}^{+}$, if $k^{2}<a<(k+1)^{2}$, let $a=k^{2}+m, m=1,2, \cdots$, $2 k, \sqrt{a}=k+\theta, 0<\theta<1$, then
$$
\left[\frac{1}{\{\sqrt{a}\}}\right]=\left[\frac{1}{\sqrt{a}-k}\right]=\left[\frac{\sqrt{a}+k}{a-k^{2}}\right]=\left[\frac{2 k+\theta}{m}\right],
$$

Since $0<\frac{2 k+\theta}{m}-\frac{2 k}{m}<1$, if there exists an integer $t$ between $\frac{2 k}{m}$ and $\frac{2 k+\theta}{m}$, then $\frac{2 k}{m}<t<\frac{2 k+\theta}{m}$. Thus, on one hand, $2 k<m t$, so $2 k+1 \leqslant m t$, and on the other hand,
$$
m t<2 k+\theta<2 k+1 \text {, a contradiction. Hence }\left[\frac{2 k+\theta}{m}\right]=\left[\frac{2 k}{m}\right],
$$

Therefore,
$$
\begin{array}{l}
\sum_{k^{2}<a<(k+1)^{2}}\left[\frac{1}{\{a\}}\right]=\sum_{m=1}^{2 k}\left[\frac{2 k}{m}\right], \\
\sum_{n=1}^{(m+1)^{2}} f(a)=\sum_{k=1}^{n} \sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right] .
\end{array}
$$

Next, we calculate $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$: Draw a $2 k \times 2 k$ table, in the $i$-th row, place a “ $\star$ ” at every multiple of $i$, then this row has $\left[\frac{2 k}{i}\right]$ “ $\star$ ”s. The total number of “ $\star$ ”s in the entire table is $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$. On the other hand, count the “ $\star$ ”s by column; in the $j$-th column, if $j$ has $T(j)$ positive divisors, then this column has $T(j)$ “ $\star$ ”s. Therefore, the total number of “ $\star$ ”s in the entire table is $\sum_{j=1}^{2 k} T(j)$, so $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]=\sum_{j=1}^{2 k} T(j)$.
Example:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$i$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline 1 & * & * & $\star$ & $\star$ & * & * \\
\hline 2 & & $\star$ & & $\star$ & & * \\
\hline 3 & & & * & & & * \\
\hline 4 & & & & $\star$ & & \\
\hline 5 & & & & & * & \\
\hline 6 & & & & & & $\star$ \\
\hline
\end{tabular}

Thus,
$$
\begin{aligned}
\sum_{a=1}^{(n+1)^{2}} f(a)= & \sum_{k=1}^{n} \sum_{j=1}^{2 k} T(j) \\
= & n[T(1)+T(2)]+(n-1)[T(3)+T(4)]+ \\
& \cdots+[T(2 n-1)+T(2 n)] .
\end{aligned}
$$

Therefore, $\sum_{k=1}^{16^{2}} f(k)=\sum_{k=1}^{15}(16-k)[T(2 k-1)+T(2 k)]$.
Let $a_{k}=T(2 k-1)+T(2 k), k=1,2, \cdots, 15$, it is easy to get the values of $a_{k}$ as follows:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$k$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline$a_{k}$ & 3 & 5 & 6 & 6 & 7 & 8 & 6 & 9 & 8 & 8 & 8 & 10 & 7 & 10 & 10 \\
\hline
\end{tabular}

Thus,
$$
\sum_{k=1}^{256} f(k)=\sum_{k=1}^{15}(16-k) a_{k}=783 .
$$

By definition, $f(256)=f\left(16^{2}\right)=0$, when $k \in\{241,242, \cdots, 255\}$, let $k=15^{2}+$ $r(16 \leqslant r \leqslant 30)$,
$$
\begin{array}{c}
\sqrt{k}-15=\sqrt{15^{2}+r}-15=\frac{r}{\sqrt{15^{2} \mid r}+15}, \\
\frac{r}{31}<\frac{r}{\sqrt{15^{2}+r}+15}<\frac{r}{30}, \\
1 \leqslant \frac{30}{r}<\frac{1}{\left\{\sqrt{\left.15^{2}+r\right\}}\right.}<\frac{31}{r}<2,
\end{array}
$$

Then
$$
\left[\frac{1}{\{\sqrt{k}\}}\right]=1, k \in\{241,242, \cdots, 255\},
$$

Thus,
$$
\sum_{k=1}^{220} f(k)=783-\sum_{k=1}^{256} f(k)=783-15=768 .
$$

Note: This problem can also be solved by enumeration.
(Let $[x]$ denote the greatest integer not exceeding $x$, and $\{x\}=x-[x]$). Find the value of $\sum_{k=1}^{240} f(k)$.