$\mathbf{N 2 7}$ (25-2, Netherlands) Find a pair of positive integers $a, b$, satisfying:
(1) $a b(a+b)$ is not divisible by 7;
(2) $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$.
Solve $(a+b)^{7}-a^{7}-b^{7}$
$$
\begin{array}{l}
=7 a^{6} b+21 a^{5} b^{2}+35 a^{1} b^{3}+35 a^{3} b^{4}+21 a^{2} b^{5}+7 a b^{6} \\
=7 a b\left[\left(a^{5}+b^{5}\right)+3 a b\left(a^{3}+b^{3}\right)+5 a^{2} b^{2}(a+b)\right] \\
=7 a b(a+b)\left[a^{4}+2 a^{3} b+3 a^{2} b^{2}+2 a b^{3}+b^{4}\right] \\
=7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2} .
\end{array}
$$

Since 7 divides $a b(a+b)$, $7^{7} \mid 7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2}$, so $7^{3} \mid\left(a^{2}+a b+b^{2}\right)$, which means $343 \mid\left(a^{2}+a b+b^{2}\right)$.
Taking $a=18, b=1$, then $a^{2}+a b+b^{2}=343$, and it is clearly that $7 \nmid a b(a+b)$.