Three. (20 points) Let the general term of the sequence $\left\{a_{n}\right\}$ be
$$
a_{n}=n+\frac{2006}{n^{2}}(n=1,2, \cdots) \text {. }
$$

Find the minimum term of the sequence $\left\{a_{n}\right\}$.
$$
\begin{array}{l}
\text { Let } f(x)=x+\frac{2006}{x^{2}}. \\
\text { When } 0<x<15, \text { we have } \\
2006\left(\frac{1}{15 \times 15^{2}}+\frac{1}{15^{2} \times 15}\right)=\frac{2006 \times 2}{15^{3}}>1,
\end{array}
$$

Therefore, when $0<x<15$,
$$
\begin{array}{l}
f\left(x_{2}\right)-f\left(x_{1}\right) \\
=\left(x_{2}-x_{1}\right)\left[1-\frac{2006\left(x_{1}+x_{2}\right)}{x_{1}^{2} x_{2}^{2}}\right] . \\
\text { And } \frac{2006\left(x_{1}+x_{2}\right)}{x_{1}^{2} x_{2}^{2}}=2006\left(\frac{1}{x_{1} x_{2}^{2}}+\frac{1}{x_{1}^{2} x_{2}}\right) \\
>1 \text {, so } f\left(x_{2}\right)-f\left(x_{1}\right)<0 \text {, which means } f(x) \text { is decreasing on } (0,15).
\end{array}
$$

When $x_{1} \geqslant 16$ and $x_{2} \geqslant 16$,
$$
\begin{array}{l}
f\left(x_{2}\right)-f\left(x_{1}\right) \\
=\left(x_{2}-x_{1}\right)\left[1-\frac{2006\left(x_{1}+x_{2}\right)}{x_{1}^{2} x_{2}^{2}}\right] . \\
\text { And } \frac{2006\left(x_{1}+x_{2}\right)}{x_{1}^{2} x_{2}^{2}}=2006\left(\frac{1}{x_{1} x_{2}^{2}}+\frac{1}{x_{1}^{2} x_{2}}\right) \\
<1 \text {, so } f\left(x_{2}\right)-f\left(x_{1}\right)>0 \text {, which means } f(x) \text { is increasing on } [16,+\infty).
\end{array}
$$

Thus, the minimum term of the sequence $\left\{a_{n}\right\}$ is
$$
\min \left\{a_{15}, a_{16}\right\}=\min \left\{\frac{5381}{225}, \frac{6102}{256}\right\}=\frac{6102}{256} \text { . }
$$

Therefore, the minimum term of the sequence $\left\{a_{n}\right\}$ is $a_{16}=\frac{6102}{256}$.