25. Given the first 10 terms of the sequence $\left\{a_{n}\right\}$ are
$$
\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \text {. }
$$

Then, the 2006th term of this sequence is $\qquad$, and the sum of the first 2006 terms is $\qquad$.
25. $\frac{53}{64}, 998 \frac{55}{64}$.

For the sequence, group it as follows:
$$
\left(\frac{1}{2}\right),\left(\frac{1}{3}, \frac{2}{3}\right),\left(\frac{1}{4}, \frac{2}{4}, \frac{3}{4}\right),\left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right), \cdots \text {. }
$$

Each group has 1 term, 2 terms, 3 terms, 4 terms, $\qquad$
Let the denominator of the 2006th term be $n$, then
$$
1+2+\cdots+(n-1) \geqslant 2006>1+2+\cdots+(n-2) \text {. }
$$

Since $n$ is a positive integer, so $n=64$.
Thus, the 2006th term is in the 63rd group, and the first 62 groups have $\frac{62 \times 63}{2}=1953$ terms.
Since $2006-1953=53$, the 2006th term is $\frac{53}{64}$.
Summing each group yields a new sequence:
$$
\frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \cdots, \frac{61}{2}, 31, \frac{1+2+\cdots+53}{64} \text {. }
$$

The sum of the first 2006 terms is
$$
\begin{array}{l} 
\frac{1}{2}+1+\frac{3}{2}+2+\frac{5}{2}+3+\cdots+\frac{61}{2}+31+\frac{1+2+\cdots+53}{64} \\
=(1+2+\cdots+31)+\left(\frac{1}{2}+\frac{3}{2}+\cdots+\frac{61}{2}\right)+ \\
\frac{1+2+\cdots+53}{64} \\
= 496+480 \frac{1}{2}+\frac{1431}{64}=998 \frac{55}{64} .
\end{array}
$$
(Provided by Hong Zhibao)