This is just combinatorics.

Condition (c) is equivalent to $a+c=b.$ For a given value of $b,$ there are exactly $b-1$ pairs of positive integers $(a,c)$ summing to $b,$ but if $b$ is even, we have to throw out $a=c=b/2,$ so there are $b-2$ solutions for even $b$ and $b-1$ solutions for odd $b.$ So the total number of solutions is $0+0+2+2+4+4+\cdots+1996+1996+1998 = 4(1+2+\cdots+998)+1998 = 998\cdot 999 \cdot 2 + 1998 = 2 \cdot 999^2 = \fbox{1996002}.$