Consider the number $\underbrace{55\dots5}_{n \text{ times}}$. The sum of the squares of its digits is $n \cdot 5^2 = 25n$. We can exchange any two fives by one three and one four, so the sum of the squares decreases by $5^2$, until we run out of fives. So we can get any sum from $25 \cdot \lfloor n/2 \rfloor$ and $25 \cdot n$. So it suffices to show that there is an integer $k$ such that $\frac{n}{2} \le k^2 \le n$. Choose $k$ such that $k^2 \le n < (k+1)^2$. Suppose $k^2 < \frac{n}{2}$. Then $n > 2k^2$, and $(k+1)^2 > n > 2k^2 \implies (k+1)^2 \ge 2k^2+2 \iff k^2 - 2k + 1 \le 0 \iff (k-1)^2 \le 0$, which is false except for $k=1$, or $2 < n < 4$, that is, $n=3$. But the statement of the problem itself gives an example with $n$ digits: $122$.