i found that the answer is $5$ it was so easy am i right?

Indeed it's easy an generalizes to an $n \times n$ grid for $n \ge 3$ to get $5$ as the highest possible sum. Originally I got tripped up because I thought you couldn't use $0$, but after I realized you could it was easy. To show $5$ is attainable, put $1$ into $(2,1), (2,2), (2,3), (1,2), (3,2)$ and then $0$ into anything else. Now we show $5$ is sharp. Suppose a sum of $S$ is attainable. Then it follows that some column has sum at least $S-2$ because let the sum of the $m^{th}$ column be $C_m$. Then as soon as $C_1 + C_2 + ... + C_k > 1$, we need $C_1 + C_2 + ... + C_k \ge S-1$ obviously. Then by similar arguments that column has an element that is at least $S-4$, but this means $S-4 \le 1 \implies S \le 5$ so we're done.

This could be generalized to an n-dimensional analog; the answer there is $2n+1$. Same idea with dinoboy's proof

Note that a sum of $5$ may be achieved in the following configuration: [asy][asy] /* APMO 2012 Problem 2, free script by liberator, 1 November 2014 */ unitsize(0.8cm); defaultpen(fontsize(16pt)); /* Draw objects */ draw(Line((0,0), (0,3), 0.2, 0)); draw(Line((1,0), (1,3), 0.2, 0)); draw(Line((2,0), (2,3), 0.2, 0)); draw(Line((3,0), (3,3), 0.2, 0)); draw(Line((0,0), (3,0), 0, 0.2)); draw(Line((0,1), (3,1), 0, 0.2)); draw(Line((0,2), (3,2), 0, 0.2)); draw(Line((0,3), (3,3), 0, 0.2)); /* Label objects */ label("$0$", (0,0), 1.8*dir(45)); label("$1$", (0,1), 1.8*dir(45)); label("$0$", (0,2), 1.8*dir(45)); label("$1$", (1,0), 1.8*dir(45)); label("$1$", (1,1), 1.8*dir(45)); label("$1$", (1,2), 1.8*dir(45)); label("$0$", (2,0), 1.8*dir(45)); label("$1$", (2,1), 1.8*dir(45)); label("$0$", (2,2), 1.8*dir(45)); label("$\vdots$", (1,0), 1.8*dir(-45)); label("$\cdots$", (3,1), 1.8*dir(45)); [/asy][/asy] where each other square has entry zero. We show that $5$ is the maximum. Suppose that a value of $k$ is attainable. We work on a general $n \times n$ board. Choose \[\ell = \max_{1 \leq j \leq n} \left \{ j: \sum_{i=1}^{j-1} C_i \leq 1 \right \},\] where $C_i$ denotes the sum of the entries in column $i$. Now choose \[ \ell' = \min_{\ell \leq j \leq n} \left \{j: \sum_{i= j+1}^n C_i \leq 1 \right \}.\] If $\ell < \ell'$, then we have from definition \[\sum_{i=1}^{\ell} C_i > 1, \sum_{i= \ell +1}^n C_i > 1,\] which contradicts the problem requirement. Hence $\ell = \ell'$. Similarly, if $R_i$ denotes the sum of the entries in row $i$, $\exists$ an integer $1 \leq m \leq n$ such that \[\sum_{i=1}^{m-1} R_i \leq 1, \sum_{i= m+1}^n R_i \leq 1.\] Considering the value of the entry in column $\ell$, row $m$, we have \[1 \geq k- \left (\sum_{i=1}^{\ell-1} C_i + \sum_{i= \ell +1}^n C_i + \sum_{i=1}^{m-1} R_i + \sum_{i= m+1}^n R_i \right ) \geq k - 4.\] Therefore $k \leq 5$; in particular, this holds in the case $n=2012$, as required.

Let $d_i$ be the sum of $i$-th column , $c_i$ be the sum of $i$-th row Suppose $d_1+..+d_k < 1 , d_1+..+d_{k+1} > 1$ , then $k-th$ column divides the grid into 2 parts$ S_1 , S_2$ with $S_1 \leq 1 , S_2 \leq 1$ Similary $l$-th row divides the grid into 2 parts $T_1,T_2$ with the same property . Let $A$ be the number at the intersection of these row and column Let $X,Y$ be the sum of the numbers in the $k$-th column from above and under the square has number $A$ Suppose that $d_k > 3$ , because $0 \leq A \leq 1$ then $X+Y > 2$ , then suppose $X > 1$ then $T_1 > 1$ contradiction . then , $d_k \geq 3$ so the sum of all the numbers is at most $5$

we claim that the answer is $5$ the construction for 5 is easy like above first define a square/column to be arabic if the sum of the squares/columns before it is less than 1 but if add this square/column will be larger than 1 note that the sum of squares\columns after the arabic one won't be larger that $1$ obsevation: there's no column with sum more that $3$ proof: suppose the contrary then start from the top of that column and sum up the squares until we rach at the arabic square $a$ we know that $|a| \le 1$ thus the sum of squares after $a$ will be large than $1$ and this is a contradiction Now start from the most left row and start sum the column until we reach at an arabic column this column has at most sum 3 but columns before it have at most sum 1 and the same for those after it so the whole sum is smaller than $5$ and we are done

Can someone explain these solutions? I'm kinda confused.

The answer is $\boxed{5}$. A quick Solution and Motivation: probably the most instructive way a problem can teach you IVT and extremal simultaneously (both of which I think are algebraic techniques, in a combinatorial problem). $\color{green} \rule{25cm}{2pt}$ $\color{green} \textbf{Value Control.}$ Define an $\textit{intersection cell}$ of the configuration to be a cell $I_C$ so that the sum of rows above it, rows under it, columns at the left of it, and columns at the right of it are each at most $1$. Then, the box exhibits an $\textit{intersection cell}$. $\color{green} \rule{25cm}{0.4pt}$ $\color{green} \textbf{Proof.}$ Consider a row $r$ where the sum of the first $r$ rows first exceed $1$, and consider a column $c$ similarly. We will prove that the cell $I_C$ in row $r$, column $c$ is an intersection cell. However, this is incredibly easy (coming up with this Claim is the hard part); for the extremal condition require the top $r-1$ rows' sum to be $1$ or less, and the leftmost $c-1$ columns similarly. Moreover, the problem condition require the bottom $2012-r$ rows to have a sum of at most $1$, since the topmost $r$ rows already exceed $1$. The rightmost $2012-c$ columns are analogous. $\blacksquare$ $\blacksquare$ $\color{red} \rule{25cm}{2pt}$ $\color{red} \textbf{Finishing Touches.}$ Let the sum of the squares which is strictly at the top-left of $I_C$ to be $a$, directly at the top to be $b$, strictly top-right $c$, directly left $d$, exactly $I_C$ $e$, directly right $f$, strictly bottom-left $g$, directly bottom $h$, strictly bottom-right $i$. Then, $a+b+c \leq 1$, $a+d+g \leq 1$, $c+f+i \leq 1$, $g+h+i \leq 1$ and $e \leq 1$. Summing all that, it becomes \[ 2a+b+2c+d+e+f+2g+h+2i \leq 1+1+1+1+1 = 5\]with equality (sum of $a$ to $i$ equalling five) iff $a,c,g,i$ all equals zero; $b = 1-0-0 = d = f = i$, and $e = 1$. Not only does we obtain an upper bound $(5)$, but also a configuration which equality happens and $\textit{all}$ possible configurations which equality happens. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$

Denote by $D_i$, the sum of all the rows below $R_i$ ($i^{th}$ row)(rows are counted from top to bottom)(define $U_i$ similarly for above) and by $L_i$, the sum of all the columns on the left of $C_i$ ($i^{th}$ column)(columns are counted from left to right)(define $E_i$ similarly for the right). Also, $a_{i,j}$ is the entry in the cell formed by the intersection of $R_i$ and $C_j$. Let $R_x$ be such that $U_x \leq 1$ & $U_{x+1} > 1$. And let $C_y$ be such that $L_y \leq 1$ & $L_{y+1} > 1$ [If no such $R_i$ ($C_j$) exists then x=n (y=n)... this case will be explored later] So, due to the problem condition, $D_x \leq 1$ & $E_y \leq 1$ Now let $S_U(i,j)$ be the sum of all the entries in $C_j$ above the cell (i,j) (define $S_D(i,j)$ for below) and let $S_L(i,j)$ be the sum of all the entries in $R_i$ and on the left of the cell (i,j) (define $S_E(i,j)$ for the right). Since, only non-negative entries fill the grid, $U_x \leq 1 \implies S_U(x,y) \leq 1$ and $D_x \leq 1 \implies S_D(x,y) \leq 1$. Hence, Sum of all Numbers = $L_y + E_y + S_U(x,y) + S_D(x,y) + a_{x,y} \leq 1+1+1+1+ a_{x,y} \leq 4+ a_{x,y} \leq 5$ ( and $x,y \in \{1, 2, 3, ..., n-1\}$ ) where equality holds iff $a_{x,y}=S_U(x,y)=S_D(x,y)=S_L(x,y)=S_E(x,y)=1$ and all other entries are 0. $\dashrightarrow$ ( $\ast$ ) Finally, if exactly one of x, y is n, WLOG x=n then $S_D(x,y)=0$ and the bound reduces to 4. And similarly, if x=y=n, the bound reduces to 3. Hence, the maximum possible sum must be $\textbf{5}$ for which all possible equality configurations have already been found above. ( $\ast$ )

Here is a solution which I believe is clean. (-er than all solutions given above)

Cute problem. We show that the answer is five. For construction take $3*3$ board and fill the corner and middle squares with a $1$ and the rest with a $0$. So ,assume the contrary. Now, if we look at the middlemost line, dividing it into $2012*1011$ rectangles. One of them has sum at most $1$ while the other has sum at least $4$ . Thereafter keep adding rows to the one with sum at most $1$, until it becomes the rectangle with sum at least $4$. Thus the row that was just added had sum at least $3$.If the addition of rows, to the smaller block,doesn't affect it,then we get that the last row has sum at least $4$. Both of these cases are handled similarly. If we examine the desired row,. we show that we can partition the row ,into two parts,such that each of them has sum at least $1$. Just start from the left and move to the right till the sum of the elements is between $(1,2]$ , which is certainly doable as all the cells in it can atmost have a $1$ in them. However ,we extend the line which partitions the row to the entire board,we are done.

I have a cool construction solution. if we have an NxN grid, we can just circle the topmost, bottommost, leftmost, and rightmost Nx(floor of (N-1)/2) grids, as anything else with a grid value less than floor((N-1)/2) will definitely be less than or equal to the value for the Nx(floor of (N-1)/2) grid. Then the maximum value would be 4-(intersectional regions. The max would then be the intersectional regions =0. Then the leftover 1x1 or 2x2 square has a max value of 1. So the answer is 5

I claim the answer is $5$. First, I claim there exists a row such that the sum of everything to the strict left and strict right are both at most $1$. If splitting the bottommost row gives that above is $\leq 1$, then we are done. Similarly, if splitting the topmost row gives that below is $\leq 1$, then are are done. Otherwise, there exists a row in the middle travelling downward such that everything strictly above that row is $\leq 1$ and everything strictly below is $\leq 1$. By symmetry, the same applies to columns. Let $K$ be the square in this row and this column. Now, strict left + strict right + strict up + strict down + K counts everything in the grid at least once. Thus this gives an upper bound on the overall sum, and this is bounded above by $1 + 1 + 1 + 1 +1 = $. The construction is putting $1$ in $(1006, 1006), (1, 1006), (1006, 1), (2012, 1006), (1006, 2012)$ and $0$ everywhere else.