One. (20 points) Arrange natural numbers in a triangular number table as shown in Figure 4, following the principle of increasing from top to bottom and from left to right (each row has one more number than the previous row). Let $a_{i j}\left(i, j \in \mathbf{N}_{+}\right)$ be the number located in the $j$-th position from left to right in the $i$-th row from the top of this triangular number table (for example, $a_{42}=8$).
(1) If $a_{i j}=2008$, find the values of $i$ and $j$.
(2) Let the sum of the numbers in the $n$-th row from the top of the triangular number table be $b_{n}$, and define
$$
c_{n}=\left\{\begin{array}{ll}
1, & n=1 ; \\
\frac{n}{b_{n}-n}, & n \geqslant 2 .
\end{array}\right.
$$

If the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$ is $T_{n}$, find $T_{n}$.
(1) In the triangular number table, the first $n$ rows contain
$$
1+2+\cdots+n=\frac{n(n+1)}{2}
$$

numbers, i.e., the last number in the $i$-th row is $\frac{i(i+1)}{2}$.
Therefore, the $i$ that makes $a_{i j}=2.008$ is the smallest positive integer solution to the inequality $\frac{i(i+1)}{2} \geqslant 2008$.

Since $\frac{62 \times 63}{2}=1953$, and $\frac{63 \times 64}{2}=2016$, we have $i=63$. Thus, the first number in the 63rd row is $\frac{62 \times 63}{2}+1=1954$.
Hence, $j=(2008-1954)+1=55$.
(2) The sum of all natural numbers in the first $n$ rows is
$$
\begin{array}{l}
S_{n}=\frac{1}{2} \times \frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+1\right] \\
=\frac{n(n+1)\left(n^{2}+n+2\right)}{8} .
\end{array}
$$

Then $b_{n}=S_{n}-S_{n-1}=\frac{n\left(n^{2}+1\right)}{2}$.
Therefore, when $n \geqslant 2$,
$$
\begin{aligned}
c_{n}= & \frac{n}{b_{n}-n}=\frac{2}{n^{2}-1}=\frac{1}{n-1}-\frac{1}{n+1}, \\
T_{n}= & 1+\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+ \\
& \left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n+1}\right) \\
= & 1+1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}=\frac{5}{2}-\frac{1}{n}-\frac{1}{n+1} .
\end{aligned}
$$