9. It is known that the number of $t$-digit numbers is
$$m=\underbrace{88 \cdots 89}_{(t-1) \uparrow} \text { (numbers). }$$

Let the $m$ $t$-digit numbers in $A$ be $a_{m}$, $a_{m-1}, \cdots, a_{1}$, i.e.,
$$A=\sum_{i=1}^{m} a_{i} \times 10^{t(i-1)} .$$

Thus, $A \equiv \sum_{i=1}^{m} a_{i}(\bmod 9)$.
Let $t \equiv r(\bmod 9)(2 \leqslant r \leqslant 8)$. Then
$m=\underbrace{88 \cdots 89}_{(t-1) \uparrow} \equiv 8(t-1)+9$
$\equiv 10-r(\bmod 9)$.
Since $\sum_{i=k}^{k+8} i=\frac{9(2 k+8)}{2} \equiv 0(\bmod 9)$, we have
$A \equiv \sum_{i=0}^{9-1}(\underset{i \uparrow}{99 \cdots 9}-i) \equiv \sum_{i=0}^{9-1}(9-i)=\sum_{j=r}^{9} j$
$=\frac{(10-r)(9+r)}{2}=\frac{r(1-r)}{2}(\bmod 9)$.
When $r=2,5,8$, $A \equiv-1(\bmod 9)$;
When $r=3,7$, $A \equiv-3(\bmod 9)$;
When $r=4,6$, $A \equiv 3(\bmod 9)$.
If $A \equiv 3$ or $-3(\bmod 9)$, then $A$ is a multiple of 3, but not a multiple of 9; if $A \equiv-1(\bmod 9)$, then $A \equiv-1(\bmod 3)$, but $n^{2} \equiv 0$ or $1(\bmod 3)$.
Thus, $A$ is not a perfect square.