6. $N$ is a perfect square, all of its digits are less than 7, and when each digit is increased by 3, it remains a perfect square. Then the value of $N$ is $\qquad$ .
6. $N=1$, or 16, or 1156, or $\stackrel{k \uparrow}{11} \stackrel{k-1 \uparrow}{155 \cdots 56}$.

Let $N=a \cdot 10^{3}+b \cdot 10^{2}+c \cdot 10+d$,
$a, b, c, d$ are natural numbers, and $1 \leqslant a \leqslant b, 0 \leqslant b, c, d \leqslant 6$, by the problem $N=n^{2}$ ( $n$ is some positive integer), so $n^{2}=N \leqslant$ 6666, hence $n \leqslant 81$. Also,
$(a+3) 10^{3}+(b+3) 10^{2}+(c+3) 10+(d+3)=m^{2}(m$ is some positive integer $)$, $m^{2}-n^{2}=333,(m+n)(m-n)=3 \cdot 11 \cdot 101$.
Since $m+n>m-n$, and $n \leqslant 81$, thus $m+n=101$ and $m-n=33$.
Therefore, $n=34$, so $N=1156$.