$1 \cdot 159$ Find the positive integer pairs $a, b$, such that:
(1) $a b(a+b)$ is not divisible by 7;
(2) $(a+b)^{7}-a^{7}-b^{7}$ is divisible by $7^{7}$.

Verify your answer.
［Solution］We have
$$(a+b)^{7}-a^{7}-b^{7}=7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2}$$

From (1) and (2), we know that $7^{7} \mid 7 a b(a+b)\left(a^{2}+a b+b^{2}\right)^{2}$, which means $7^{3} \mid a^{2}+a b+b^{2}$.

Next, we need to find the positive integer pairs $(a, b)$ that satisfy
$$(a+b)^{2}-a b=a^{2}+a b+b^{2}=7^{3}=343$$

Since $0<a b \leqslant \frac{1}{4}(a+b)^{2}$, we can derive the inequality from (1):
$$343 \leqslant(a+b)^{2} \leqslant 457$$

Solving this, we get $19 \leqslant a+b \leqslant 21$.
Let $a+b=19$, from (1) we get $a b=18$, solving these equations simultaneously gives $a=18, b=1$. It is easy to verify that this is a pair of positive integers that satisfy (1) and (2).