For each positive integer $n$, define the function $f(n)=\left\{\begin{array}{ll}0 & \text { when } n \text { is a perfect square, } \\ {\left[\frac{1}{\{\sqrt{n}\}}\right]} & \text { when } n \text { is not a perfect square. }\end{array}\right.$ (where $[x]$ denotes the greatest integer not exceeding $x$, and $\{x\}=x-[x])$, find: $\sum_{k=1}^{240} f(k)$.

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The function $f(n)$ is defined as follows:
- $f(n) = 0$ if $n$ is a perfect square.
- $f(n) = \left[\frac{1}{\{\sqrt{n}\}}\right]$ if $n$ is not a perfect square.

Here, $[x]$ represents the greatest integer less than or equal to $x$, and $\{x\}$ represents the fractional part of $x$, which is $x - [x]$.

We need to find the value of $\sum_{k=1}^{240} f(k)$.
For any $a, k \in \mathbf{N}^{+}$, if $k^{2}<a<(k+1)^{2}$, then $1 \leqslant a-k^{2} \leqslant 2 k$. Let $\sqrt{a}=k+\theta, 0<\theta<1$, then $\frac{1}{\sqrt{a}}=\frac{1}{\theta}=\frac{1}{\sqrt{a}-k}=\frac{\sqrt{a}+k}{a-k^{2}}=\frac{2 k+\theta}{a-k^{2}}<\frac{2 k}{a-k^{2}}+1, \therefore\left[\frac{1}{\sqrt{a}}\right]=\left[\frac{2 k}{a-k^{2}}\right]$. Let $a$ run through all integers in the interval $\left(k^{2},(k+1)^{2}\right)$, then $\sum_{k^{2}<a<(k+1)^{2}}\left[\frac{1}{\{a\}}\right]=\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$, thus $\sum_{a=1}^{(n+1)^{2}} f(a)=\sum_{k=1}^{n} \sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$. (1) 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 “*” at every multiple of $i$, then the number of “*” in this row is $\left[\frac{2 k}{i}\right]$, and the total number of “*” in the entire table is $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$; on the other hand, collect the “*” by column: in the $j$-th column, if $j$ has $T(j)$ positive divisors, then this column has $T(j)$ “*”, so the total number of “*” in the entire table is $\sum_{j=1}^{2 k} T(j)$, thus $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]=\sum_{j=1}^{2 k} T(j)$. An example is as follows:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline 1 & * & * & * & * & * & * \\
\hline 2 & & * & & * & E & * \\
\hline 3 & & & * & & & * \\
\hline 4 & & & & * & & \\
\hline 5 & & & & & * & $=$ \\
\hline 6 & & & & & & * \\
\hline
\end{tabular}

Then $\sum_{a=1}^{(n+1)^{2}} f(a)=\sum_{k=1}^{n} \sum_{j=1}^{2 k} T(j)=n[T(1)+T(2)]+(n-1)[T(3)+T(4)]+\cdots+[T(2 n-1)+T(2 n)]$, (2)
Thus, $\sum_{k=1}^{16^{2}} f(k)=\sum_{k=1}^{15}(16-k)[T(2 k-1)+T(2 k)]$,

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}

Therefore, $\sum_{k=1}^{256} f(k)=\sum_{k=1}^{15}(16-k) a_{k}=783$.
By definition, $f(256)=f\left(16^{2}\right)=0$, and for $k \in\{241, 242, \cdots, 255\}$, let $k=15^{2}+r \quad(16 \leqslant r \leqslant 30)$,
$$
\begin{array}{l}
\sqrt{k}-15=\sqrt{15^{2}+r}-15=\frac{r}{\sqrt{15^{2}+r}+15}, \frac{r}{31}<\frac{r}{\sqrt{15^{2}+r}+15}<\frac{r}{30}, 1 \leqslant \frac{30}{r}<\frac{1}{\left\{\sqrt{15^{2}+r}\right\}} \\
<\frac{31}{r}<2, \\
\quad \text { then }\left[\frac{1}{\{\sqrt{k}\}}\right]=1, k \in\{241,242, \cdots, 255\} \text { (5), thus } \sum_{k=1}^{240} f(k)=783-\sum_{k=1}^{256} f(k)=783-15=768 .
\end{array}
$$