3 (Full score: 50 points) For each positive integer $n$, define the function
$$
f(n)=\left\{\begin{array}{ll}
0, & \text { when } n \text { is a perfect square, } \\
{\left[\frac{1}{\{\sqrt{n}\}}\right],} & \text { when } n \text { is not a perfect square. }
\end{array}\right.
$$
(Here $[x]$ denotes the greatest integer not exceeding $x$, and $\{x\}=x-[x]$). Find the value of: $\sum_{k=1}^{240} f(k)$.
For any $a, k \in \mathbf{N}_{+}$, if $k^{2}<a<(k+1)^{2}$, set
$$
\begin{array}{c}
a=k^{2}+m, m=1,2, \cdots, 2 k, \\
\sqrt{a}=k+\theta, 0<\theta<1,
\end{array}
$$

then
$$
\left[\frac{1}{\{\sqrt{a}\}}\right]=\left[\frac{1}{\sqrt{a}-k}\right]=\left[\frac{\sqrt{a}+k}{a-k^{2}}\right]=\left[\frac{2 k+\theta}{m}\right],
$$

because
$$
0<\frac{2 k+\theta}{m}-\frac{2 k}{m}<1,
$$

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 $2 k+1 \leqslant m t$, on the other hand,
$$
m t<2 k+\theta<2 k+1,
$$

which is a contradiction. Hence
$$
\left[\frac{2 k+\theta}{m}\right]=\left[\frac{2 k}{m}\right],
$$

so
$$
\sum_{k=2<(k+1)^{2}}\left[\frac{1}{\{a\}}\right]=\sum_{m=1}^{2 k}\left[\frac{2 k}{m}\right] \text {, }
$$

thus
$$
\sum_{a=1}^{(n+1)^{2}} f(a)=\sum_{k=1}^{n} \sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right] \text {. }
$$
20 points
Next, 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, fill in “ * ” at every multiple of $i$, then the number of “ * ” in this row is $\left[\frac{2 k}{i}\right]$, the total number of “ * ” in the entire table is $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$; on the other hand, count the number of “ * ” 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 $\sum_{j=1}^{2 k} T(j)$, therefore $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]=\sum_{j=1}^{2 k} T(j)$.
Example:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline j & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline 1 & $*$ & $*$ & $*$ & $*$ & $*$ & $*$ \\
\hline 2 & & $*$ & & $*$ & & $*$ \\
\hline 3 & & & $*$ & & & $*$ \\
\hline 4 & & & & $*$ & & \\
\hline 5 & & & & & $*$ & \\
\hline 6 & & & & & & $*$ \\
\hline
\end{tabular}

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

thus $\sum_{k=1}^{16^{2}} f(k)=\sum_{k=1}^{15}(16-k)[T(2 k-1)+T(2 k)]$. . .30 points

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
$$
\sum_{k=1}^{256} f(k)=\sum_{k=1}^{15}(16-k) a_{k}=783 .
$$
$\qquad$
By definition $f(256)=f\left(16^{2}\right)=0$, when $k \in\{241,242, \cdots, 255\}$, set
$$
k=15^{2}+r(16 \leqslant r \leqslant 30),
$$
$$
\begin{array}{c}
\sqrt{k}-15=\sqrt{15^{2}+r}-15=\frac{r}{\sqrt{15^{2}+r}+15}, \\
\frac{r}{31}<\frac{r}{\sqrt{15^{2}+r}+15}<\frac{r}{30}, \\
1 \leqslant \frac{30}{r}<\frac{1}{\left.\sqrt{15^{2}+r}\right\}}<\frac{31}{r}<2,
\end{array}
$$

then $\left[\frac{1}{\{\sqrt{k}\}}\right]=1, k \in\{241,242, \cdots, 255\}$,
thus
$$
\sum_{k=1}^{200} f(k)=783-\sum_{k=1}^{255} f(k)=783-15=768 .
$$

Note: This problem can also be solved by enumeration.