jmc

It is well-known that $\tau (n)$ is odd if and only if $n$ is a perfect square. (Otherwise, we can group divisors into pairs whose product is $n$.) Thus, $S(n)$ is odd if and only if there are an odd number of perfect squares less than $n$. So $S(1),S(2)$ and $S(3)$ are odd, while $S(4),S(5),\dots ,S(8)$ are even, and $S(9),\dots ,S(15)$ are odd, and so on. So, for a given $n$, if we choose the positive integer $m$ such that $m^2\le n<(m+1{)}^2$ we see that $S(n)$ has the same parity as $m$. It follows that the numbers between $1^2$ and $2^2$, between $3^2$ and $4^2$, and so on, all the way up to the numbers between $43^2$ and $44^2=1936$ have $S(n)$ odd. These are the only such numbers less than $2005$ (because $45^2=2025>2005$). Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are $3$ numbers between $1$ (inclusive) and $4$ (exclusive), $5$ numbers between $4$ and $9$, and so on. The number of numbers from $n^2$ to $(n+1{)}^2$ is $(n+1-n)(n+1+n)=2n+1$. Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus, $a=[2(1)+1]+[2(3)+1]\dots [2(43)+1]=3+7+11\dots 87$. $b=[2(2)+1]+[2(4)+1]\dots [2(42)+1]+70=5+9\dots 85+70$, the $70$ accounting for the difference between $2005$ and $44^2=1936$, inclusive. Notice that if we align the two and subtract, we get that each difference is equal to $2$. Thus, the solution is $|a-b|=|b-a|=|2\cdot 21+70-87|=\boxed{25}$.