Is there a way to express the block matrix with 1 in the position (i,j) of the block [i,j] and 0 otherwise in some compact form?

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$\delta_{[i,j]}$ is defined as the $N^2\times N^2$ block matrix whose elements $(i,j)$ in the block $[i,j]$ are equal to 1, and all the others are 0.

I'm looking for a way to express it as a function of appropriates $N\times N$ matrices, for example as a kronecker product or sum.

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$ \def\bbR#1{{\mathbb R}^{#1}} \def\e{{\bf e}} \def\B{{\bf B}} \def\E{{\bf E}} \def\J{{\bf J}} \def\n{\nabla}\def\o{{\tt1}}\def\p{\partial} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\qif{\quad\iff\quad} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\c#1{\color{red}{#1}} $Let $I\in\bbR{n\times n}$ denote the identity matrix and let $\e_k$ denote its $k^{th}$ column vector.

The $\e_k$ vectors are the standard vector basis for $\bbR{n}.\:$ Using the $\e_k$ vectors, an arbitrary matrix $A\in\bbR{n\times n}$ can be decomposed into its constituent scalar elements $$\eqalign{ A_{kl} &= \e_k^TA\e_l \qif A = \sum_{k=1}^n\sum_{l=1}^n \e_kA_{kl}\e_l^T \\ }$$ Block-analogs of the $\e_k$ vectors can be constructed using the Kronecker product $$\eqalign{ \E_k &= \e_k\otimes I \\ }$$ It should come as no surprise that the $\E_k$ matrices can be used to decompose a block matrix $\B\in\bbR{n^2\times n^2}$ into its constituent blocks $B_{kl}\in\bbR{n\times n}$ $$\eqalign{ B_{kl} &= \E_k^T\B\E_l \qif \B = \sum_{k=1}^n\sum_{l=1}^n \E_kB_{kl}\E_l^T \\ }$$ Let $J\in\bbR{n\times n}$ denote the all-ones matrix. Then setting $B_{ij}=J$ and all other $B_{kl}=0$, the summation reduces to a single non-zero term and the desired block matrix is $${\E_i J \E_j^T}$$