Let $\phi(n)$ denote $\textit{Euler's phi function}$, the number of integers $1\leq i\leq n$ that are relatively prime to $n$.  (For example, $\phi(6)=2$ and $\phi(10)=4$.)  Let \[S=\sum_{d|2008}\phi(d),\] in which $d$ ranges through all positive divisors of $2008$, including $1$ and $2008$.  Find the remainder when $S$ is divided by $1000$.
1. **Understanding Euler's Totient Function**: Euler's totient function, denoted as $\phi(n)$, counts the number of integers from $1$ to $n$ that are relatively prime to $n$. For example, $\phi(6) = 2$ because the numbers $1$ and $5$ are relatively prime to $6$.

2. **Given Problem**: We need to find the sum of $\phi(d)$ for all divisors $d$ of $2008$, and then find the remainder when this sum is divided by $1000$.

3. **Key Theorem**: The sum of Euler's totient function over all divisors of $n$ is equal to $n$. Mathematically, this is expressed as:
   \[
   \sum_{d|n} \phi(d) = n
   \]
   This theorem can be proven by considering the fractions $\frac{1}{n}, \frac{2}{n}, \ldots, \frac{n}{n}$ and noting that there are exactly $\phi(d)$ fractions with denominator $d$ for each divisor $d$ of $n$.

4. **Application to the Problem**: Applying the theorem to $n = 2008$, we get:
   \[
   \sum_{d|2008} \phi(d) = 2008
   \]

5. **Finding the Remainder**: We need to find the remainder when $2008$ is divided by $1000$. This can be done using the modulo operation:
   \[
   2008 \mod 1000 = 8
   \]

Conclusion:
\[
\boxed{8}
\]