4. Let $N=23 x+92 y$ be a perfect square, and $N$ does not exceed 2,392. Then the number of all positive integer pairs $(x, y)$ that satisfy the above conditions is $\qquad$ pairs.


4.27.
$\because N=23 x+92 y=23(x+4 y)$, and 23 is a prime number, $N$ is a perfect square not exceeding 2392,
$\therefore x+4 y=23 m^{2}$ ( $m$ is a positive integer) and
$N=23^{2} \cdot m^{2} \leqslant 2392$,
thus $m^{2} \leqslant \frac{2392}{23^{2}}=\frac{104}{23}<5$.
Solving, we get $m^{2}=1$ or 4.
When $m^{2}=1$, from $x+4 y=23$, we get $y=1,2,3,4$, $5, x=19,15,11,7,3$;

When $m^{2}=4$, from $x+4 y=92$, we get $y=1,2,3,4$, $5, \cdots, 22, x=88,84,80, \cdots, 4$.

Therefore, there are $(1,19),(2,15),(3,11),(4,7),(5,3)$ and $(1,88),(2,84), \cdots,(22,4)$.
Hence, the number of pairs $(x, y)$ that satisfy the conditions is $5+22=27$.