Let $b$ be a positive integer. Determine all $2002$-tuples of non-negative integers $(a_{1}, a_{2}, \cdots, a_{2002})$ satisfying \[\sum^{2002}_{j=1}{a_{j}}^{a_{j}}=2002{b}^{b}.\]
1. Let \( c = \max_j a_j \). Then we have \( c^c \leq S = \sum_{j=1}^{2002} a_j^{a_j} \leq 2002 c^c \).

2. If \( c \geq 737 \), then we need to show that \( 2002(c-1)^{c-1} < c^c \). This inequality can be verified by considering the growth rates of the functions involved. Specifically, for large \( c \), \( c^c \) grows much faster than \( 2002(c-1)^{c-1} \).

3. Given \( c \geq 737 \), we have \( b \geq c \). This implies that \( a_j = c \) for all \( j \). Therefore, the only solution in this case is \( a_j = c \) for all \( j \).

4. When \( b < 737 \), we need to consider other possible solutions. For example, if \( b = 1 \), then \( 2002b^b = 2002 \). This can be achieved by setting \( a_1 = a_2 = \cdots = a_k = 1 \) and \( a_{k+1} = \cdots = a_{2002} = 0 \), where \( k = 2002 \).

5. To generalize, for any \( b \), we can set \( a_1 = a_2 = \cdots = a_b = b \) and \( a_{b+1} = \cdots = a_{2002} = 0 \). This satisfies the equation \(\sum_{j=1}^{2002} a_j^{a_j} = b \cdot b^b = 2002b^b\).

The final answer is \( \boxed{ (a_1, a_2, \cdots, a_{2002}) } \) such that \(a_1 = a_2 = \cdots = a_b = b\) and \(a_{b+1} = \cdots = a_{2002} = 0\).