Four, (Full marks 20 points) For what rational values of $x$ is the algebraic expression $9 x^{2}+23 x-2$ the product of two consecutive positive integers?

When $x$ is what rational number, the algebraic expression $9 x^{2}+23 x-2$ is the product of two consecutive positive integers?
Let two consecutive positive even numbers be $k, k+2$. Then we have
$$
9 x^{2}+23 x-2=k(k+2) \text {, }
$$

which simplifies to $9 x^{2}+23 x-\left(2+k^{2}+2 k\right)=0$. Since $x$ is a rational number, the discriminant must be a perfect square, i.e.,
$$
\begin{array}{l} 
\Delta=23^{2}+4 \times 9\left(k^{2}+2 k+1+1\right) \\
=565+[6(k+1)]^{2} . \\
\text { Let } \Delta=p^{2} \text { (where } p \geq 0\text{). Then, } \\
p^{2}-[6(k+1)]^{2}=565 \\
=113 \times 5=565 \times 1, \\
{[p+6(k+1)][p-6(k+1)] } \\
=113 \times 5=565 \times 1 .
\end{array}
$$

Since $p \geqslant 0$ and $k>0$, we get
$$
\left\{\begin{array}{l}
p+6(k+1)=113 . \\
p-6(k+1)=5 .
\end{array}\right.
$$

Solving this, we find $k=8$. Substituting $k=8$ into the original equation, we get $x=2$ or $x=-\frac{41}{9}$; or
$$
\left\{\begin{array}{l}
p+6(k+1)=565, \\
p-6(k+1)=1 .
\end{array}\right.
$$

Solving this, we find $k=46$. Substituting $k=46$ into the original equation, we get
$$
x=-17 \text { or } x=\frac{130}{9} \text {. }
$$

In summary, when $x=2$ or $x=-17$, $9 x^{2}+23 x-2$ is exactly the product of the two positive even numbers 8 and 10, or 46 and 48.
Note: When $k=46$, substituting into the original equation, we get
$$
\begin{array}{l}
9 x^{2}+23 x-2210=0 . \\
\Delta=80089 . \\
\because 280^{2}=78400,290^{2}=84100, \text { and the last digit of } 80089 \text { is } 9, \\
\therefore 80089 \text { can only be } 283^{2} \text { or } 287^{2}. \\
\text { After calculation, we find } 283^{2}=80089.
\end{array}
$$