At least two of the nonnegative real numbers $ a_1,a_2,...,a_n$ aer nonzero. Decide whether $ a$ or $ b$ is larger if

$ a\equal{}\sqrt[2002]{a_1^{2002}\plus{}a_2^{2002}\plus{}\ldots\plus{}a_n^{2002}}$ 

and

$ b\equal{}\sqrt[2003]{a_1^{2003}\plus{}a_2^{2003}\plus{}\ldots\plus{}a_n^{2003} }$
1. We are given two expressions for \(a\) and \(b\):
   \[
   a = \sqrt[2002]{a_1^{2002} + a_2^{2002} + \ldots + a_n^{2002}}
   \]
   \[
   b = \sqrt[2003]{a_1^{2003} + a_2^{2003} + \ldots + a_n^{2003}}
   \]

2. To compare \(a\) and \(b\), we can rewrite these expressions in a more general form. Let \(k\) be a positive integer, then:
   \[
   a_k = \sqrt[k]{a_1^k + a_2^k + \ldots + a_n^k}
   \]

3. We need to compare \(a_{2002}\) and \(a_{2003}\). Notice that:
   \[
   a_{2002} = \sqrt[2002]{a_1^{2002} + a_2^{2002} + \ldots + a_n^{2002}}
   \]
   \[
   a_{2003} = \sqrt[2003]{a_1^{2003} + a_2^{2003} + \ldots + a_n^{2003}}
   \]

4. Consider the function \(f(k) = \sqrt[k]{a_1^k + a_2^k + \ldots + a_n^k}\). We need to show that \(f(k)\) is a decreasing function of \(k\).

5. To do this, we can use the fact that for any nonnegative real numbers \(x_1, x_2, \ldots, x_n\) and for \(k < m\), the following inequality holds:
   \[
   \left( \frac{x_1^k + x_2^k + \ldots + x_n^k}{n} \right)^{\frac{1}{k}} \geq \left( \frac{x_1^m + x_2^m + \ldots + x_n^m}{n} \right)^{\frac{1}{m}}
   \]
   This is a consequence of the Power Mean Inequality.

6. Applying this to our case, we have:
   \[
   \left( \frac{a_1^{2002} + a_2^{2002} + \ldots + a_n^{2002}}{n} \right)^{\frac{1}{2002}} \geq \left( \frac{a_1^{2003} + a_2^{2003} + \ldots + a_n^{2003}}{n} \right)^{\frac{1}{2003}}
   \]

7. Since \(a_1, a_2, \ldots, a_n\) are nonnegative real numbers and at least two of them are nonzero, the sums \(a_1^{2002} + a_2^{2002} + \ldots + a_n^{2002}\) and \(a_1^{2003} + a_2^{2003} + \ldots + a_n^{2003}\) are positive. Therefore, we can remove the normalization by \(n\) and still maintain the inequality:
   \[
   \left( a_1^{2002} + a_2^{2002} + \ldots + a_n^{2002} \right)^{\frac{1}{2002}} \geq \left( a_1^{2003} + a_2^{2003} + \ldots + a_n^{2003} \right)^{\frac{1}{2003}}
   \]

8. This implies:
   \[
   a \geq b
   \]

9. Since the function \(f(k)\) is strictly decreasing, we have:
   \[
   a > b
   \]

The final answer is \( \boxed{ a > b } \)