Let the first row have $n$ children and total rows be $r$. Then the numbers per row form an AP with difference $-3$ and sum $630$:

$\displaystyle \frac{r}{2}\big(2n-3(r-1)\big)=630$ $ \;\Rightarrow\; 1260=r\big(2n-3(r-1)\big)$.

Hence $r\mid1260$. Also the last row must be positive: $n-3(r-1)>0$.

Testing the options (and ensuring $n$ is integer and last term positive):

• $r=4$: $1260/4=315$, $2n=315+3\cdot3=324 \Rightarrow n=162$, last term $=162-9=153>0$ ✓
• $r=5$: $1260/5=252$, $2n=252+3\cdot4=264 \Rightarrow n=132$, last term $=132-12=120>0$ ✓
• $r=6$: $1260/6=210$, $2n=210+3\cdot5=225 \Rightarrow n=112.5$ (not integer) ✗

Therefore, the impossible number of rows is $\boxed{6}$.