The answer is $\boxed{\textit{no}}$. Suppose otherwise, that there exists such $n$. Suppose $n$ has $k$ digits. Then, we know that \[ 10^k >n \ge \underbrace{666 \cdots 6}_{k \ \text{times}} \]Now, if $n = \underbrace{666 \cdots 6}_{k \ \text{times}}$, then $n^2 \equiv 6 \ (\text{mod} \ 10)$, and hence $n^2$ ends in $6$, a contradiction. So, we get a tighter bound \[ 10^k > n > \frac{2}{3} \cdot 10^k \]Therefore, we conclude that \[ 10^{2k} > n^2 > \frac{4}{9} \cdot 10^{2k} > \frac{4}{9} \cdot (10^{2k} - 1) \]This forces \[ \underbrace{999 \cdots 9}_{2k \ \text{times}} = 10^{2k} - 1 \ge n^2 > \frac{4}{9} (10^{2k} - 1) = \underbrace{444 \cdots 4}_{2k \ \text{times}} \]which forces a contradiction as $n^2$ can't have all its digits less than $5$.

Proposed by Nazar Agakhanov

Metropolises 2020 P2 wrote: Does there exist a positive integer $n$ such that all its digits (in the decimal system) are greather than 5, while all the digits of $n^2$ are less than 5? The answer is no. Assume that such a $n$ existed. Suppose that $n$ has $M$ digits. Then, since all of its digits are $\geq 6$ we obtain $$10^M>n \geq \underbrace{666 \cdots 6}_{M \ \text{times}}=\frac{2(10^M-1)}{3}$$ Note now that $10^{2M}>n^2>10^{2M-2}$, that is $n^2$ has either $2M-1$ or $2M$ digits. Assume firstly that it has $2M-1$ digits. Then, $$ \dfrac{4(10^M-1)^2}{9} \leq n^2 \leq 10^{2M-1} $$hence $9 \cdot 10^{2M-1} \geq 4(10^M-1)^2$ which fails due to size reasons. Hence $n^2$ must have $2M$ digits. Combining this with the fact that all its digits are $\leq 4$ we conclude that $$\dfrac{4(10^{2M-1})}{9} \geq n^2 \geq \dfrac{4(10^M-1)^2}{9}$$hence $$(2 \cdot 10^M)^2>(3n)^2 \geq (2 \cdot 10^M-2)^2$$which implies that $3n \in \{2 \cdot 10^M-2, 2 \cdot 10^M-1 \}$. The second choice is impossible due to $\pmod 3$ reasons, hence we must have $n=\dfrac{2(10^M-1)}{3}$. That is, $n=\underbrace{666 \cdots 6}_{M \ \text{times}}$ which is a contradiction since $n^2 \equiv 6 \pmod 10$, therefore its last digit would be $6$.

The answer is no. Suppose such number exists, call it $A$ CLAIM. All digits of that number must be $6$. Proof. Suppose $10^{k-1}<A<10^{k}$, then if some digits of that number is greater than $6$, we must have $A>\frac{2}{3}\cdot 10^k$ and so $$A^2>\frac{4}{9}10^{2k}>\overline{44...4}$$contradiction. $\blacksquare$ But then obviously the unit digit of $A^2$ will be $6$, contradiction.