Solution1:
\[ \begin{aligned} &\sum_{i=1}^n\sum_{i=1}^nijgcd(i,j) \ =&\sum_{d=1}^dd\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}ijd^2[\ gcd(i, j)=1\ ]\ =&\sum_{d=1}^nd^3\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}ij\sum_{k|(i,j)}\mu(k)\ =&\sum_{d=1}^{n}d^3\sum_{k=1}^{\frac{n}{d}}\mu(k)\sum_{i=1}^{\frac{n}{dk}}\sum_{i=1}^{\frac{n}{dk}}ijk^2\ =&\sum_{d=1}^{n}d^3\sum_{k=1}^{\frac{n}{d}}\mu(k)k^2(1+2+3+...+\frac{n}{dk})^2\ =&\sum_{T=1}^{n}(\frac{(1+T)T}{2})^2\sum_{d|T}d^3(\frac{T}{d})^2\mu(\frac{T}{d})\ \ \ \ \ \ \ (ps:T=dk)\ =&\sum_{T=1}^{n}(\frac{(1+T)T}{2})^2T^2\sum_{d|T}d\mu(\frac{T}{d})\ =&\sum_{T=1}^{n}(\frac{(1+T)T}{2})^2T^2\varphi(T)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ps:Id* \mu=\varphi) \end{aligned} \]\[ \begin{aligned} &\sum_{i=1}
