Find all pairs of natural numbers $(k, m)$ such that for any natural $n{}$ the product\[(n+m)(n+2m)\cdots(n+km)\]is divisible by $k!{}$.

[i]Proposed by P. Kozhevnikov[/i]
To solve the problem, we need to find all pairs of natural numbers \((k, m)\) such that for any natural number \(n\), the product \((n+m)(n+2m)\cdots(n+km)\) is divisible by \(k!\).

1. **Prime Factorization and \(\nu_p\) Notation**:
   - Let \(p\) be a prime number.
   - The \(\nu_p(x)\) notation represents the highest power of \(p\) that divides \(x\).

2. **Divisibility Condition**:
   - We need to show that \(\nu_p((n+m)(n+2m)\cdots(n+km)) \geq \nu_p(k!)\) for all primes \(p\).

3. **Case 1: \((k!, m) = 1\)**:
   - If \((k!, m) = 1\), then \(m\) is not divisible by any prime factor of \(k!\).
   - Consider a prime \(p\) that divides \(k\). Since \((p, m) = 1\), \(m\) is not divisible by \(p\).

4. **Prime Power Divisibility**:
   - We need to show that for any prime \(p\) dividing \(k\), the product \((n+m)(n+2m)\cdots(n+km)\) has at least as many factors of \(p\) as \(k!\).
   - The number of times \(p\) appears in \(k!\) is given by:
     \[
     \nu_p(k!) = \sum_{i=1}^{\infty} \left\lfloor \frac{k}{p^i} \right\rfloor
     \]
   - Among any \(p^i\) consecutive numbers, there is at least one multiple of \(p^i\). Therefore, in the sequence \((n+m), (n+2m), \ldots, (n+km)\), there are at least \(\left\lfloor \frac{k}{p^i} \right\rfloor\) multiples of \(p^i\).

5. **Verification**:
   - Since \((k!, m) = 1\), for any prime \(p\) dividing \(k\), the product \((n+m)(n+2m)\cdots(n+km)\) will have at least \(\nu_p(k!)\) factors of \(p\).
   - Thus, \(\nu_p((n+m)(n+2m)\cdots(n+km)) \geq \nu_p(k!)\).

6. **Case 2: \(p \leq k\) divides \(m\)**:
   - If \(p \leq k\) divides \(m\), choose \(n\) such that \((n, p) = 1\).
   - In this case, \(n + im\) for \(i = 1, 2, \ldots, k\) will not necessarily be divisible by \(p\), and thus the product \((n+m)(n+2m)\cdots(n+km)\) may not be divisible by \(k!\).

Conclusion:
- The pairs \((k, m)\) such that \((k!, m) = 1\) satisfy the condition.
- If \(p \leq k\) divides \(m\), the condition is not satisfied.

The final answer is \( \boxed{ (k, m) } \) such that \((k!, m) = 1\).