Answer: 25

Small Hint:

$\tau(n)$ is odd exactly when $n$ is a perfect square, so $S(n)$ changes parity only at squares

Big Hint:

$S(n)$ is odd exactly when $\lfloor\sqrt{n}\rfloor$ is odd; count $n \le 2005$ block by block between consecutive squares

Solution:

Divisors of $n$ pair up as $d$ and $\frac{n}{d},$ so $\tau(n)$ is odd exactly when $n$ is a perfect square. Hence $S(n)$ changes parity exactly at the squares, which means $S(n)$ is odd exactly when the number of squares up to $n,$ namely $\lfloor\sqrt{n}\rfloor,$ is odd.

For each $k,$ there are $2k + 1$ integers $n$ with $\lfloor\sqrt{n}\rfloor = k,$ namely $k^2 \le n \le k^2 + 2k.$ Since $44^2 = 1936$ $\le 2005$ $\lt 2025 = 45^2,$ the odd values $k = 1, 3, \ldots, 43$ all have their full blocks within range, so $$\begin{aligned} a &= \sum_{k \text{ odd},\, k \le 43} (2k + 1) \\ &= 2(1 + 3 + \cdots + 43) + 22 \\ &= 2 \cdot 484 + 22 \\ &= 990. \end{aligned}$$

Then $b = 2005 - 990 = 1015,$ and $|a - b| = 25.$