I have the following double $\sum_{i=[1,\ldots,n], j=[1,\ldots,m]: i\neq j }\frac{1}{j}$.
If I ignore restriction $i \neq j$ then $$ \sum_{i=[1,\ldots,n], j=[1,\ldots,m] }\frac{1}{j}=n \sum_{ j=[1,\ldots,m] }\frac{1}{j} \le n(1+\ln(m)) $$
Now, I am not sure how to it with this restiction?
With the restriction you have
$$\begin{align*}\sum_{j=1}^m\;\sum_{i\in[1..n],i\ne j}\frac1j&=\begin{cases} \sum_{j=1}^m\frac{n-1}j,&\text{if }n\ge m\\\\ \sum_{j=1}^m\frac{n-1}j+\sum_{j=n+1}^m\frac1j,&\text{if }n<m \end{cases}\\\\ &=\begin{cases} (n-1)H_m,&\text{if }n\ge m\\\\ (n-1)H_m+(H_m-H_n),&\text{if }n<m \end{cases}\\\\ &=\begin{cases} (n-1)H_m,&\text{if }n\ge m\\\\ nH_m-H_n,&\text{if }n<m\;, \end{cases} \end{align*}$$
where $H_n$ is the $n$-the harmonic number.