Find one pair of positive integers $a,b$ such that $ab(a+b)$ is not divisible by $7$, but $(a+b)^7-a^7-b^7$ is divisible by $7^7$.
1. **Expression Simplification**:
   We start with the given expression:
   \[
   (a+b)^7 - a^7 - b^7
   \]
   Using the binomial theorem, we can expand \((a+b)^7\):
   \[
   (a+b)^7 = \sum_{k=0}^{7} \binom{7}{k} a^{7-k} b^k
   \]
   Subtracting \(a^7\) and \(b^7\) from this expansion, we get:
   \[
   (a+b)^7 - a^7 - b^7 = \sum_{k=1}^{6} \binom{7}{k} a^{7-k} b^k
   \]

2. **Factoring the Expression**:
   We can factor the expression as follows:
   \[
   (a+b)^7 - a^7 - b^7 = 7ab(a+b)(a^2 + ab + b^2)^2
   \]
   This factorization can be verified by expanding the right-hand side and comparing terms.

3. **Divisibility Condition**:
   We need to ensure that:
   \[
   7^7 \mid 7ab(a+b)(a^2 + ab + b^2)^2
   \]
   This simplifies to:
   \[
   7^6 \mid ab(a+b)(a^2 + ab + b^2)^2
   \]
   Since \(7^6 = 7 \cdot 7^5\), we need:
   \[
   7^5 \mid ab(a+b)(a^2 + ab + b^2)^2
   \]

4. **Finding \(a\) and \(b\)**:
   We need to find \(a\) and \(b\) such that:
   \[
   7 \nmid ab(a+b)
   \]
   and:
   \[
   7^3 \mid a^2 + ab + b^2
   \]
   Let's test \(a = 18\) and \(b = 1\):
   \[
   a^2 + ab + b^2 = 18^2 + 18 \cdot 1 + 1^2 = 324 + 18 + 1 = 343 = 7^3
   \]
   This satisfies the condition \(7^3 \mid a^2 + ab + b^2\).

5. **Checking \(ab(a+b)\)**:
   We need to ensure that \(7 \nmid ab(a+b)\):
   \[
   ab(a+b) = 18 \cdot 1 \cdot (18 + 1) = 18 \cdot 19 = 342
   \]
   Since \(342\) is not divisible by \(7\), this condition is also satisfied.

Conclusion:
The pair \((a, b) = (18, 1)\) satisfies both conditions.

The final answer is \(\boxed{(18, 1)}\)