There's something I don't understand: if we cut out the corners, then the top and bottom rows have only $n$ squares left each, and the left and right columns have only $m$ squares left each, so we should write $2(m+n)$ numbers instead of $2(m+n-2)$. Am I right?

Oh,excuse me you are right It's $2(m+n)$

Couldn't we just write the system of equations and work with determinants and stuff?

May,be but I did'nt do this

I thought for a moment and I think I've got a solution (It is inspired by another problem we did in IMO training last week in Oberwolfach, a kind of 3D functional equation): Consider the matrix A sending every element to the difference of itself by the arithmetic mean of its four neighbours. If A is surjective, we have proved even more than we wanted to but in any case, we're done. But A is surjective if it is injective(since it operates on a finite-dimensional vector space). But A is injective iff its kernel consists only of o, so let's show that there is only one way to choose elements when we write zeros at the first and last rows and columns. So, let x be a vector for which Ax=o and take the element of x with greatest value, call it x0. Every element of x is the arithmetic mean of its four neighboring elements, so either one of them has greater value than x0, in contradiction to the choice of x0, or all are equal to x0. Continuing, we get that all elements of the vector x are equal, so must be zero. Thus, we have proved that A has nonzero discriminant. This even shows that the choice of the elements is unique. btw, I believe there is a constructive proof and think I heard a talk about the continuous version of that problem, but I don't really remember. Sorry if my English is too bad, but I'm not that trained in writing about mathematics in English, sorry Peter

Proof: Consider the continuous function $f: [a,b]^{mn}\rightarrow[a,b]^{mn}$ which maps each element of table to average of it 4 neighbors.($b$ is maximum of $2(m+n)$ numbers and $a$ is minimum of $2(m+n)$ numbers.) Obviously $f$ is continuous. So by Brower's fixed point theorem it has a fixed point. Fixed point is a table satisfying problem conditions.

Here is an approach via random walks. Given an interior point $p$, consider a random walk starting at $p$, and continuing until it hits the boundary at some point $q$. Let $f(p)$ be the expected value of $f(q)$. Then $f$ has the averaging property almost by definition.

Another proof: We must prove that if we put all numbers around the table zero, then the only case for the table is that all elements are zero. (That proves that the matrix is invertible and everything will be proved.) Suppose in this case we have filled the table with numbers - not all zero - fulfilling the conditions. Suppose the maximum number in the table is $A$, then all of its neighbors must be equal to $A$. Doing this again for its neighbors you see that $A=0$. So the matrix of equations is invertible and always has solution.

Another Solution: Fill the table with arbitrary numbers, and at each step put each number to average of its 4 neighbors. We could prove with some analytic work that the table will converge to a table satisfying the conditions.

Omid Hatami wrote: Proof: Consider the continuous function $f: [a,b]^{mn}\rightarrow[a,b]^{mn}$ which maps each element of table to average of it 4 neighbors.($b$ is maximum of $2(m+n)$ numbers and $a$ is minimum of $2(m+n)$ numbers.) Obviously $f$ is continuous. So by Brower's fixed point theorem it has a fixed point. Fixed point is a table satisfying problem conditions. that's cooooool!