Let $n_1,n_2,\ldots,n_s$ be distinct integers such that
$$(n_1+k)(n_2+k)\cdots(n_s+k)$$is an integral multiple of $n_1n_2\cdots n_s$ for every integer $k$. For each of the following assertions give a proof or a counterexample:

$(\text a)$ $|n_i|=1$ for some $i$
$(\text b)$ If further all $n_i$ are positive, then
$$\{n_1,n_2,\ldots,n_2\}=\{1,2,\ldots,s\}.$$
1. **Initial Observation**: Note that none of the \( n_i \) equals 0, since taking \( k \) large gives \( \prod_{i=1}^s (n_i + k) > 0 \), hence not a multiple of 0.

### Part (a)
2. **Assumption for Contradiction**: Suppose \( |n_i| \ge 2 \) for all \( i \). Let \( N := \prod_{i=1}^s |n_i| \).

3. **Divisibility Conditions**: Given that \( N \) divides \( \prod_{i=1}^s |n_i + k| \) for every integer \( k \), we specifically have:
   \[
   N \mid \prod_{i=1}^s |n_i + 1|
   \]
   and
   \[
   N \mid \prod_{i=1}^s |n_i - 1|
   \]

4. **Multiplying the Conditions**: Multiplying these two conditions, we get:
   \[
   N^2 \mid \prod_{i=1}^s |(n_i + 1)(n_i - 1)| = \prod_{i=1}^s |n_i^2 - 1|
   \]

5. **Inequality**: Since \( N = \prod_{i=1}^s |n_i| \), we have:
   \[
   N^2 = \left( \prod_{i=1}^s |n_i| \right)^2 = \prod_{i=1}^s n_i^2
   \]
   Therefore, the divisibility condition implies:
   \[
   \prod_{i=1}^s n_i^2 \leq \prod_{i=1}^s (n_i^2 - 1)
   \]

6. **Contradiction**: Since \( |n_i| \ge 2 \), we have \( n_i^2 - 1 < n_i^2 \) for each \( i \). Thus:
   \[
   \prod_{i=1}^s (n_i^2 - 1) < \prod_{i=1}^s n_i^2
   \]
   This is a contradiction. Hence, our assumption that \( |n_i| \ge 2 \) for all \( i \) is false. Therefore, \( |n_i| = 1 \) for some \( i \).

### Part (b)
7. **Function Definition**: Let \( f(x) = \frac{(x + n_1) \cdots (x + n_s)}{N} \), where \( N = n_1 n_2 \cdots n_s \). It is given that \( f: \mathbb{Z} \rightarrow \mathbb{Z} \).

8. **Polynomial Expansion**: We can expand \( f(x) \) as:
   \[
   f(x) = \sum_{k=0}^s c_k \binom{x}{k}
   \]
   where \( c_k \in \mathbb{Z} \) for all \( k \).

9. **Leading Term**: The leading term of \( f(x) \) is:
   \[
   \frac{s!}{N} \binom{x}{s}
   \]
   Since \( f(x) \) maps integers to integers, \( \frac{s!}{N} \) must be an integer. Hence, \( N \mid s! \).

10. **Distinct Positive Integers**: Since \( N \) is a product of the \( s \) distinct positive integers \( n_i \), and \( N \mid s! \), the only way this can happen is if \( \{n_1, n_2, \ldots, n_s\} = \{1, 2, \ldots, s\} \).

The final answer is \( \boxed{ \{n_1, n_2, \ldots, n_s\} = \{1, 2, \ldots, s\} } \)