Bin packing problem or not?
You can't simply say: "This is a bin packing problem and therefore NP-hard."
Consider the problem of $N$ unit bins and $M$ unit items with the target of maximizing the usage of the bins. Clearly this is also a bin packing problem, but it is not NP-hard.

However, you can reduce to the decision variant of your problem from partition:


  Given a set of items with sizes $a_1, \dotsc, a_k$, can it be partitioned into two subsets of equal size.


Given items $a_1, \dotsc, a_k$, we construct an instance of your problem as follows:


Let $A_i = \sum_{j=1}^i a_j$ amd $A=A_k$.
We introduce $2A+2$ items and $2A+2$ bins.
$g_{i,j} = \begin{cases}
A-a_l &amp; A_{l-1}&lt;i=j\leq A_l\\
A-a_l &amp; A+A_{l-1}&lt;i=j\leq A+A_l\\
0 &amp; A_{l-1}&lt;i,j\leq A_l,~ i\neq j\\
0 &amp; A+A_{l-1}&lt;i,j\leq A+A_l,~ i\neq j\\
2 &amp; A_{l-1}&lt;i\leq A_l,~ A+A_{l-1}&lt;j\leq A+A_l\\
2 &amp; A+A_{l-1}&lt;i\leq A+A_l,~ A_{l-1}&lt;j\leq A_l\\
0 &amp; j&gt;2A,~ i\neq j\\
\frac A2 &amp; i=j&gt;2A\\
0 &amp; i=2A+1,~ j\leq A\\
0 &amp; i=2A+2,~ A&lt;j\leq 2A\\
1 &amp; \text{otherwise}
\end{cases}$
We ask for an assignment with $\sum_{i=1}^{A}\sum_{j=1}^{A} x_{i,j} \geq A+2$.


The construction works as follows:


Any solution will satisfy $x_{i,j}=1 \Rightarrow i=j$. 
Items/bins $i$ with $A_{l-1} &lt; i \leq A_l$ resp. $A+A_{l-1} &lt; i \leq A+A_l$ correspond to $a_l$. In a solution, exactly one of the groups will be taken, corresponding to $a_l$ being in the first resp. second set of the partition.
The threshold can only be reached, if the last two items are placed, ensuring that the partition is into two sets of size $\frac A2$.