(a) Does there exist a polynomial $ P(x)$ with coefficients in integers, such that $ P(d) \equal{} \frac{2008}{d}$ holds for all positive divisors of $ 2008$?
(b) For which positive integers $ n$ does a polynomial $ P(x)$ with coefficients in integers exists, such that $ P(d) \equal{} \frac{n}{d}$ holds for all positive divisors of $ n$?
Let's break down the problem into two parts and provide a detailed solution for each.

### Part (a)
We need to determine if there exists a polynomial \( P(x) \) with integer coefficients such that \( P(d) = \frac{2008}{d} \) for all positive divisors \( d \) of 2008.

1. **Identify the divisors of 2008:**
   \[
   2008 = 2^3 \times 251
   \]
   The positive divisors of 2008 are \( 1, 2, 4, 8, 251, 502, 1004, 2008 \).

2. **Set up the condition for the polynomial:**
   For each divisor \( d \) of 2008, we need:
   \[
   P(d) = \frac{2008}{d}
   \]

3. **Check the consistency of the polynomial:**
   Consider two distinct divisors \( a \) and \( b \) of 2008. We must have:
   \[
   \frac{P(a) - P(b)}{a - b} \in \mathbb{Z}
   \]
   Substituting \( P(a) = \frac{2008}{a} \) and \( P(b) = \frac{2008}{b} \), we get:
   \[
   \frac{\frac{2008}{a} - \frac{2008}{b}}{a - b} = \frac{2008(b - a)}{ab(a - b)} = \frac{2008}{ab} \in \mathbb{Z}
   \]
   This implies that for any distinct divisors \( a \) and \( b \) of 2008, \( ab \) must also be a divisor of 2008.

4. **Check if \( ab \) is a divisor of 2008 for all pairs of divisors:**
   - For \( a = 1 \) and \( b = 2 \), \( ab = 2 \), which is a divisor.
   - For \( a = 1 \) and \( b = 4 \), \( ab = 4 \), which is a divisor.
   - For \( a = 2 \) and \( b = 4 \), \( ab = 8 \), which is a divisor.
   - For \( a = 2 \) and \( b = 251 \), \( ab = 502 \), which is a divisor.
   - For \( a = 4 \) and \( b = 251 \), \( ab = 1004 \), which is a divisor.
   - For \( a = 8 \) and \( b = 251 \), \( ab = 2008 \), which is a divisor.

   Since \( ab \) is not always a divisor of 2008 for all pairs of divisors, there is no polynomial \( P(x) \) with integer coefficients that satisfies \( P(d) = \frac{2008}{d} \) for all positive divisors \( d \) of 2008.

### Part (b)
We need to determine for which positive integers \( n \) there exists a polynomial \( P(x) \) with integer coefficients such that \( P(d) = \frac{n}{d} \) for all positive divisors \( d \) of \( n \).

1. **Condition for the polynomial:**
   For any distinct divisors \( a \) and \( b \) of \( n \), we must have:
   \[
   \frac{P(a) - P(b)}{a - b} \in \mathbb{Z}
   \]
   Substituting \( P(a) = \frac{n}{a} \) and \( P(b) = \frac{n}{b} \), we get:
   \[
   \frac{\frac{n}{a} - \frac{n}{b}}{a - b} = \frac{n(b - a)}{ab(a - b)} = \frac{n}{ab} \in \mathbb{Z}
   \]
   This implies that for any distinct divisors \( a \) and \( b \) of \( n \), \( ab \) must also be a divisor of \( n \).

2. **Determine the values of \( n \):**
   - If \( n \) is a prime number, its only divisors are 1 and \( n \). For these values, \( ab = n \), which is a divisor of \( n \).
   - If \( n = 1 \), its only divisor is 1, and the condition is trivially satisfied.

   Therefore, \( n \) must be either 1 or a prime number.

3. **Construct the polynomial for \( n = 1 \) or a prime number:**
   - For \( n = 1 \), the polynomial \( P(x) = 1 \) satisfies \( P(1) = 1 \).
   - For a prime number \( p \), the polynomial \( P(x) = -x + p + 1 \) satisfies:
     \[
     P(1) = -1 + p + 1 = p \quad \text{and} \quad P(p) = -p + p + 1 = 1
     \]
     Thus, \( P(d) = \frac{p}{d} \) for \( d = 1 \) and \( d = p \).

Conclusion:
\(\blacksquare\)

The final answer is \( \boxed{ n = 1 } \) or \( n \) is a prime number.