Let $b$ be a positive integer. Find all $2002$−tuples $(a_1, a_2,\ldots , a_{2002})$, of natural numbers such that
\[\sum_{j=1}^{2002} a_j^{a_j}=2002b^b.\]
1. We start with the given equation:
   \[
   \sum_{j=1}^{2002} a_j^{a_j} = 2002b^b
   \]
   where \( b \) is a positive integer and \( a_j \) are natural numbers.

2. Consider the case where all \( a_j \) are equal to \( b \). Then:
   \[
   a_j = b \quad \text{for all} \quad j = 1, 2, \ldots, 2002
   \]
   Substituting \( a_j = b \) into the sum, we get:
   \[
   \sum_{j=1}^{2002} b^b = 2002 \cdot b^b
   \]
   This matches the right-hand side of the given equation:
   \[
   2002b^b = 2002b^b
   \]
   Therefore, one solution is:
   \[
   (a_1, a_2, \ldots, a_{2002}) = (b, b, \ldots, b)
   \]

3. Next, we need to consider if there are any other possible solutions. Suppose there exists at least one \( a_j \neq b \). Without loss of generality, assume \( a_1 \neq b \). Let us analyze the implications:
   - If \( a_1 > b \), then \( a_1^{a_1} \) would be significantly larger than \( b^b \), making it difficult to balance the sum to exactly \( 2002b^b \) with the remaining terms.
   - If \( a_1 < b \), then \( a_1^{a_1} \) would be significantly smaller than \( b^b \), requiring other terms to be larger to balance the sum, which again would be difficult to achieve.

4. To formalize this, consider the growth rates of \( a_j^{a_j} \):
   - For \( a_j > b \), \( a_j^{a_j} \) grows much faster than \( b^b \).
   - For \( a_j < b \), \( a_j^{a_j} \) grows much slower than \( b^b \).

5. Given the constraints and the rapid growth of the function \( x^x \), it is highly unlikely to find a combination of different \( a_j \) values that sum to exactly \( 2002b^b \) without all \( a_j \) being equal to \( b \).

6. Therefore, the only feasible solution is:
   \[
   (a_1, a_2, \ldots, a_{2002}) = (b, b, \ldots, b)
   \]

The final answer is \( \boxed{ (a_1, a_2, \ldots, a_{2002}) = (b, b, \ldots, b) } \).