Sugerencia: Muestre que $$\sum_{l=1}^d\,\gcd(l,d)=d\,\sum_{t\mid d}\,\frac{\phi(t)}{t}$$ for every positive integer $d$, and that $$\left(\phi*\sigma_1\right)(n)=n\,\sigma_0(n)$$
para cada entero positivo $n$ donde $\phi$ es de Euler totient función, $\sigma_1$ es el divisor-función de suma, y $*$ es la convolución de Dirichlet.
Para la primera pista, contamos el número de $l\in\{1,2,\ldots,d\}$ $\gcd(l,d)=\tau$ para un determinado $\tau\mid d$. Es fácil ver que hay $\phi\left(\frac{d}{\tau}\right)$ tales enteros $l$, de donde $$\sum_{l=1}^d\,\gcd(l,d)=\sum_{\tau \mid d}\,\tau\,\phi\left(\frac{d}{\tau}\right)=d\,\sum_{t\mid d}\,\frac{\phi(t)}{t}\,.$$ Then, we get $$\begin{align}\sum_{d\mid n}\,\sum_{l=1}^d\,\gcd(l,d)&=\sum_{d\mid n}\,d\,\sum_{t\mid d}\,\frac{\phi(t)}{t}=\sum_{t\mid n}\,\phi(t)\,\sum_{\delta\mid\frac{n}{t}}\,\delta\\&=\sum_{t\mid n}\,\phi(t)\,\sigma_1\left(\frac{n}{t}\right)=\left(\phi*\sigma_1\right)(n)\,.\end{align}$$ To show the second hint, we simply verify the equality when $n$ is a prime power, as the Dirichlet convolution preserves multiplicativity. You can also show that $\text{id}=\phi*1$ and $\sigma_1=1*\text{id}$, where $\text{id}$ is the identity function $n\mapsto n$ and $1$ is the constant function $n\mapsto1$. Ergo, $$\phi*\sigma_1=\phi*(1*\text{id})=(\phi*1)*\text{id}=\text{id}*\text{id}=\text{id}\cdot\sigma_0\,.$$
