Natural numbers 1,2,3, .., 2003 are written in an arbitrary sequence $a_{1}, a_{2}, a_{3}, \ldots a_{2003}$. Let $b_{1}=1 a_{1}, b_{2}=2 a_{2}, b_{3}=3 a_{3}, \ldots, b_{2003}=2003 a_{2003}$, and $B$ be the maximum of the numbers $b_{1}, b_{2}, b_{3}, \ldots, b_{2003}$.

a) If $a_{1}=2003, a_{2}=2002, a_{3}=2001, \ldots, a_{2002}=2, a_{2003}=1$, find the value of $B$.

b) Prove that $B \geq 1002^{2}$.
a) Using the inequality between the arithmetical and geometrical mean, we obtain that $b_{n}=n(2004-n) \leq\left(\frac{n+(2004-n)}{2}\right)^{2}=1002^{2}$ for $n=1,2,3, \ldots, 2003$. The equality holds if and only if $n=2004-n$, i.e. $n=1002$. Therefore, $B=b_{1002}=1002 \times(2004-1002)=1002^{2}$. b) Let $a_{1}, a_{2}, a_{3}, \ldots a_{2003}$ be an arbitrary order of the numbers $1,2,3, \ldots, 2003$. First, we will show that numbers $1002,1003,1004, \ldots, 2003$ cannot occupy the places numbered $1,2,3$, $\ldots, 1001$ only. Indeed, we have $(2003-1002)+1=1002$ numbers and 1002 places. This means that at least one of the numbers $1002,1003,1004, \ldots, 2003$, say $a_{m}$, lies on a place which number $m$ is greater than 1001. Therefore, $B \geq m a \geq 1002 \times 1002=1002^{2}$.