Solution 1: We see that, for any positive integer $n$, $$S=\sum_{x y \mid n} \varphi(x) \cdot y=\sum_{x \mid n} \varphi(x)\left(\sum_{y \left\lvert\, \frac{n}{x}\right.} y\right)=\sum_{x \mid n} \varphi(x) \sigma\left(\frac{n}{x}\right)$$ Since $\varphi$ and $\sigma$ are both weakly multiplicative (if $x$ and $y$ are relatively prime, then $\varphi(x y)=\varphi(x) \varphi(y)$ and $\sigma(x y)=\sigma(x) \sigma(y))$, we may break this up as $$\prod_{p}(\varphi(p)+\sigma(p))$$ where the product is over all primes that divide $n$. This is simply $2^{10} n$, giving an answer of $2^{10}=1024$. Solution 2: We recall that $$\sum_{d \mid n} \varphi(d)=n$$ So, we may break up the sum as $$S=\sum_{x y \mid n} \varphi(x) \cdot y=\sum_{y \mid n} y \sum_{x \left\lvert\, \frac{n}{y}\right.} \varphi(x)=\sum_{y \mid n} y\left(\frac{n}{y}\right)$$ so $S$ is simply $n$ times the number of divisors of $n$. This number is $2^{10}=1024$. Solution 3: When constructing a term in the sum, for each prime $p$ dividing $n$, we can choose to include $p$ in $x$, or in $y$, or in neither. This gives a factor of $p-1, p$, or 1, respectively. Thus we can factor the sum as $$S=\prod_{p \mid n}(p-1+p+1)=\prod_{p \mid n} 2 p=2^{10} n$$ So the answer is 1024.