238 Test: Can 2007 be expressed in the form
$$
a_{1}^{x_{1}}+a_{2}^{x_{2}}+\cdots+a_{m}^{x_{n}}-b_{1}^{y_{1}}-b_{2}^{y_{2}}-\cdots-b_{n}^{y_{n}}
$$

where $m, n$ are both positive integers greater than 130 and less than 140 (allowing $m$ to equal $n$), $a_{1}, a_{2}, \cdots, a_{m}, b_{1}, b_{2}, \cdots, b_{n}$ are all distinct positive rational numbers less than 6, $x_{1}, x_{2}, \cdots, x_{m}, y_{1}, y_{2}, \cdots, y_{n}$ are all positive integers greater than 1 and less than 5, and $a_{1}^{x_{1}}, a_{2}^{x_{2}}, \cdots, a_{m}^{x_{n}}, b_{1}^{y_{1}}, b_{2}^{y_{2}}, \cdots, b_{n}^{y_{n}}$ are all distinct, and $a_{1}-b_{1}, a_{2}-b_{2}, \cdots, a_{t}-b_{t}, a_{1}+b_{1}, a_{2}+b_{2}, \cdots, a_{t}+b_{t} (t=\min \{m, n\})$ are all distinct. Please explain your reasoning.
Solution: It can be done. The reasons are as follows:
Let $m=n=132$,
$$
\begin{array}{l}
a_{i}=\frac{16(i+2)^{2}+(2 i+3)^{2}}{(2 i+4)(2 i+3)}(i=1,2, \cdots, 125), \\
b_{i}=\frac{16(i+2)^{2}-(2 i+3)^{2}}{(2 i+4)(2 i+3)}(i=1,2, \cdots, 125), \\
a_{k}=\frac{(2 k+4)^{2}+(2 k+3)^{2}}{2(2 k+4)(2 k+3)}(k=126,127, \cdots, 132), \\
b_{k}=\frac{(2 k+4)^{2}-(2 k+3)^{2}}{2(2 k+4)(2 k+3)}(k=126,127, \cdots, 132) .
\end{array}
$$

Obviously, for any $k \in\{1,2, \cdots, 132\}$, we have $0 < a_{i} < 6 (i=1,2, \cdots, 125)$,
$$
\begin{array}{l}
a_{k} < 6 (k=126,127, \cdots, 132),
\end{array}
$$

which is obviously true.
Thus, $a_{i}<6(i=1,2, \cdots, 132)$.
Now let $x_{1}=x_{2}=\cdots=x_{132}$
$$
=y_{1}=y_{2}=\cdots=y_{132}=2 \text {. }
$$

For $a_{i}(1 \leqslant i \leqslant 125)$, note that
$$
\begin{array}{l}
\left(2 i+3,16(i+2)^{2}+(2 i+3)^{2}\right) \\
=\left(2 i+3,16(i+2)^{2}\right) \\
=\left(2 i+3,4(i+2)^{2}\right) \\
=\left(2 i+3,(2 i+4)^{2}\right) \\
=\left(2 i+3,(2 i+3)^{2}+2(2 i+3)+1\right) \\
=(2 i+3,1)=1, \\
\left(2 i+4,16(i+2)^{2}+(2 i+3)^{2}\right) \\
=\left(2 i+4,4(2 i+4)^{2}+(2 i+4-1)^{2}\right) \\
=(2 i+4,1)=1 .
\end{array}
$$

Therefore, $a_{i}(1 \leqslant i \leqslant 125)$ is a reduced fraction.
Similarly, it can be verified that $a_{k}(126 \leqslant k \leqslant 132)$, $b_{i}(1 \leqslant i \leqslant 132)$ are all reduced fractions.

Since the denominators of $a_{1}, a_{2}, \cdots, a_{132}, b_{1}, b_{2}, \cdots, b_{132}$ are all different, then $a_{1}, a_{2}, \cdots, a_{132}, b_{1}, b_{2}, \cdots, b_{132}$ are all different. Therefore,
$$
a_{1}^{x_{1}}, a_{2}^{x_{2}}, \cdots, a_{132}^{x_{132}}, b_{1}^{y_{1}}, b_{2}^{y_{2}}, \cdots, b_{132}^{y_{132}}
$$

are also all different.
Note that
$$
\begin{array}{l}
\quad a_{i}-b_{i}=\frac{2 i+3}{i+2}, a_{i}+b_{i}=\frac{16(i+2)}{2 i+3} \\
(i=1,2, \cdots, 125), \\
\quad a_{k}-b_{k}=\frac{2 k+3}{2 k+4}, a_{k}+b_{k}=\frac{2 k+4}{2 k+3} \\
(k=126,127, \cdots, 132) .
\end{array}
$$

It is easy to verify that all these fractions are reduced fractions. Since the numerators or denominators of these reduced fractions are all different, these fractions must be all different.
Note that
$$
\begin{array}{l}
\sum_{i=1}^{125}\left(a_{i}^{2}-b_{i}^{2}\right)=\sum_{i=1}^{125}\left(a_{i}-b_{i}\right)\left(a_{i}+b_{i}\right) \\
=\sum_{i=1}^{125} 16=16 \times 125=2000, \\
\sum_{i=120}^{132}\left(a_{i}^{2}-b_{i}^{2}\right)=\sum_{i=126}^{132}\left(a_{i}-b_{i}\right)\left(a_{i}+b_{i}\right) \\
=\sum_{i=126}^{132} 1=(132-125) \times 1=7 .
\end{array}
$$

Therefore, $\sum_{i=1}^{132}\left(a_{i}^{2}-b_{i}^{2}\right)=2000+7=2007$.
In summary, 2007 can be expressed as required.