Let $f$ be a real-valued function defined on the positive integers satisfying the following condition: For all $n>1$ there exists a prime divisor $p$ of $n$ such that $f(n)=f\left(\frac{n}{p}\right)-f(p)$. Given that $f(2001)=1$, what is the value of $f(2002)$?
1. **Understanding the given condition:**
   The function \( f \) satisfies the condition that for all \( n > 1 \), there exists a prime divisor \( p \) of \( n \) such that:
   \[
   f(n) = f\left(\frac{n}{p}\right) - f(p)
   \]
   We need to find \( f(2002) \) given that \( f(2001) = 1 \).

2. **Analyzing the function for prime numbers:**
   Let \( n = p \) where \( p \) is a prime number. Then:
   \[
   f(p) = f\left(\frac{p}{p}\right) - f(p) = f(1) - f(p)
   \]
   Solving for \( f(p) \):
   \[
   2f(p) = f(1) \implies f(p) = \frac{f(1)}{2}
   \]
   Let \( f(1) = 2a \). Then:
   \[
   f(p) = a \quad \forall p \in P
   \]

3. **Analyzing the function for composite numbers:**
   Consider \( n = p_1 p_2 \) where \( p_1 \) and \( p_2 \) are prime numbers (they may be the same or different). Then:
   \[
   f(n) = f\left(\frac{n}{p_1}\right) - f(p_1) = f(p_2) - f(p_1) = a - a = 0
   \]
   Thus, \( f(n) = 0 \) for any product of two primes.

4. **Generalizing for products of more primes:**
   By induction, if \( n = p_1 p_2 p_3 \), then:
   \[
   f(n) = f\left(\frac{n}{p_1}\right) - f(p_1) = f(p_2 p_3) - f(p_1) = 0 - a = -a
   \]
   Extending this, if \( n = p_1 p_2 \cdots p_k \), then:
   \[
   f(n) = (2 - k)a
   \]

5. **Applying the function to 2001 and 2002:**
   Given \( 2001 = 3 \times 23 \times 29 \), we have:
   \[
   f(2001) = (2 - 3)a = -a
   \]
   Since \( f(2001) = 1 \), we get:
   \[
   -a = 1 \implies a = -1
   \]
   Now, for \( 2002 = 2 \times 7 \times 11 \times 13 \):
   \[
   f(2002) = (2 - 4)a = -2a = -2(-1) = 2
   \]

The final answer is \( \boxed{2} \).