$6\times 6$ grid problem [Edited to be consistent with the version I proposed in an answer below, which the OP agreed contained the essence of the problem.--John Hughes]
Some months ago, one friend proposed me a problem that I still do not find the solution. That is:
"You have a $6\times 6$ grid ($36$ squares) with an integer in each cell. There are $5$ different operators: you can add one to each cell any $n\times n$ square inside this $6\times 6$ grid (where $n=2,3,4,5,6$).
Can you, using these operators as many times as you want, make all numbers of this grid be multiples of $3$?"
The first thought it comes to me was to put all numbers modulo $3$. Then, find some coloration such that using any operator there is an invariant modulo $3$, but I did not find any. Later, I thought that using an specific combination of operators I can add some quantity to one square, so it implies that this is possible. Again, I did not find any successful combination.
I surrender, but if anyone can tell me your solution (because I do not want to tell my friend I do not solve it hehe).
 A: Assuming my interpretation of the question is correct, the answer is "no, you cannot always make the grid consist of multiples of $3$.
The proof involves a little linear algebra. We'll do everything mod 3.

*

*The space of all "problems" is a 36-dimensional space, with a basis given by the 36 matrices each of which has a "1" in exactly one location (so that $E_{3, 5}$ is all zeroes except for a $1$ in the 3rd row, fifth column).


*If we have a grid $A$, an "operation" amounts to adding a matrix $Q$ to $A$, where $Q$ is all zeroes except for a $k \times k$ block of $1$s, where $k = 2, 3, 4, 5,$ or $6$. So "solving" a puzzle $A$ amounts to finding a sequence $Q_1, Q_2, \ldots, Q_n$ where $$ A + Q_1 + Q_2 + \ldots + Q_n = 0.$$ Letting $R_i = -Q_i$, this amounts to saying $$A = R_1 + \ldots + R_n.$$


*To rephrase the previous paragraph, the puzzle is "Given any matrix $A$, is there a linear combination of the 'operation/ matrices which equals $A$?" (Note that mod 3, $-Q_1 = 2Q_i$, so each $R_i$ is in fact a linear combination of "operation" matrices.) Said even better, is the span of all operation matrices, mod 3, equal to the 36-dimensional space of `problems'?


*As it happens, I wrote a little matlab program to generate all the operation matrices, and convert each one to a 36-vector (by reading off the columns one after another). That gave me $55$ vectors in a 36 dimensional space. And then I computed the rank of this $36 \times 55$ matrix, and it turns out to be $35$ (alas).


*It takes a little thinking to see that the fact that the dimension of the column space over the reals is 35 means that the dimension of the columns space over the integers mod 3 cannot be more than 35, but that's actually true. Hence there's some "problem" matrix $A$ that cannot be represented by a linear combination (mod 3) of "operation" matrices.
A: First, consider everything mod three; second, define the "dot product" of two matrices to be
$$
A \cdot B = \sum_{i,j = 1}^3 a_{ij} b_{ij} 
$$
i.e., the result of multiplying corresponding elements and summing up the result (mod three).
Let $K$ be the matrix
 0    -1     2    -2     1     0
 1     0    -1     1     0    -1
-2     1     0     0    -1     2
 2    -1     0     0     1    -2
-1     0     1    -1     0     1
 0     1    -2     2    -1     0

which I've written with minus-signs so that it's easy to verify that for every size-2 block matrix, $X$, (i.e., a $6 \times 6$ matrix that's mostly zeroes, but with a $2 \times 2$ block of $1$s) we have
$$
K \cdot X = 0,
$$
and the same holds when $X$ is a size-3, 4, 5, or 6 block matrix.
Now suppose that the matrix $U$ given by
 0     0     0     0     0     0
 1     0     0     0     0     0
 0     0     0     0     0     0
 0     0     0     0     0     0
 0     0     0     0     0     0
 0     0     0     0     0     0

had the property that
$$
W = U + Q_1 + Q_2 + \ldots Q_n
$$
where the $Q_i$ are block matrices of sizes $2,3,4,5,$ or $6$ and $W$ is a matrix of all zeroes. Then we'd have
\begin{align}
0 &= W \cdot K \\
&= U \cdot K + Q_1 \cdot K + \ldots + Q_n \cdot K \\
&= 1  + 0  + \ldots + 0\\
& = 1,
\end{align}
which is a contradiction.
How did I come up with the matrix $K$? I looked for something whose dot product with all possible block matrices was zero; Matlab helped.
A: Clarficiation question, too long for a comment
Could the question actually be this?
Given an arbitrary $6 \times 6$ square of integers, you're allowed to perform "operations" (defined below) on the square as many times as you like. Can you, performing these operations, make all numbers be multiples of $3$, regardless of the values in the initial grid?
An "operation" consists of choosing any $2 \times 2, 3 \times 3, 4\times 4, 5 \times 5$, or $6 \times 6$ square within the grid, and adding $1$ to each of the numbers within that square.
For example, if the initial square is
$$
\begin{pmatrix}
1 & 2 & 3 &  4 & 5 & 6 \\
1 & 2 & 3 &  4 & 5 & 6 \\
1 & 2 & 3 &  4 & 5 & 6 \\
0 & 0 & 0 &  0 & 0 & 2 \\
1 & 1 & 1 &  0 & \color{red}2 & 0 \\
0 & 0 & 0 &  0 & 0 & 0 \\
\end{pmatrix}
$$
then by picking the $3 \times 3$ square centered at the red "2", we can transform this to
$$
\begin{pmatrix}
1 & 2 & 3 &  4 & 5 & 6 \\
1 & 2 & 3 &  4 & 5 & 6 \\
1 & 2 & 3 &  4 & 5 & 6 \\
0 & 0 & 0 &  \color{blue}1 & \color{blue}1 & \color{blue}3 \\
1 & 1 & 1 &  \color{blue}1 & \color{blue}3 & \color{blue}1 \\
0 & 0 & 0 &  \color{blue}1 & \color{blue}1 & \color{blue}1 \\
\end{pmatrix}
$$
where the blue entries indicate changed colors.
====
Is that a correct representation of the problem?
