Mongolian Mathematical Olympiad

Let us denote the cell in row $i$ and column $j$ by $a_{i,j}$, where $1 \leq i, j \leq 5$. Initially, $a_{i,j} = 5(i-1) + j$.

Each operation increases all numbers in a row or a column by $1$. Suppose we perform $r_i$ operations on row $i$ and $c_j$ operations on column $j$. Then the final value in cell $(i, j)$ is:

$$a_{i,j}' = a_{i,j} + r_i + c_j$$

We want all $a_{i,j}'$ to be equal. That is, for all $i, j$:

$$a_{i,j} + r_i + c_j = K$$
for some constant $K$.

Rewriting:
$$r_i + c_j = K - a_{i,j}$$

Consider two cells in the same row $i$, columns $j$ and $k$:
$$r_i + c_j = K - a_{i,j}$$
$$r_i + c_k = K - a_{i,k}$$
Subtract:
$$c_j - c_k = a_{i,k} - a_{i,j}$$
But $a_{i,k} - a_{i,j} = k - j$ (since $a_{i,j} = 5(i-1) + j$).
So $c_j - c_k = k - j$.

This means the sequence $c_1, c_2, \dots, c_5$ must satisfy $c_j - c_k = j - k$ for all $j, k$. This is only possible if $c_j = j + d$ for some constant $d$.

Similarly, consider two cells in the same column $j$, rows $i$ and $k$:
$$r_i + c_j = K - a_{i,j}$$
$$r_k + c_j = K - a_{k,j}$$
Subtract:
$$r_i - r_k = a_{k,j} - a_{i,j}$$
But $a_{k,j} - a_{i,j} = 5(k-i)$.
So $r_i - r_k = 5(k-i)$, so $r_i = 5i + e$ for some constant $e$.

Now, plug $c_j = j + d$ and $r_i = 5i + e$ into the equation:
$$r_i + c_j = K - a_{i,j}$$
But $a_{i,j} = 5(i-1) + j$, so
$$r_i + c_j = K - (5(i-1) + j)$$
$$(5i + e) + (j + d) = K - 5(i-1) - j$$
$$5i + e + j + d = K - 5i + 5 - j$$
$$5i + j + e + d + 5i + j = K + 5$$
Wait, this is inconsistent. Let's simplify:

Left: $5i + e + j + d$
Right: $K - 5(i-1) - j$

Bring all terms to one side:
$$5i + e + j + d + 5(i-1) + j - K = 0$$
But this is not matching. Let's try plugging $r_i = 5i + e$, $c_j = j + d$ into $r_i + c_j = K - a_{i,j}$:
$$5i + e + j + d = K - (5(i-1) + j)$$
$$5i + e + j + d = K - 5i + 5 - j$$
$$5i + e + j + d + 5i + j - K - 5 = 0$$
Wait, this is not matching. Let's try to solve for $K$:

Left: $5i + e + j + d$
Right: $K - 5(i-1) - j$

So:
$$5i + e + j + d = K - 5(i-1) - j$$
$$5i + e + j + d + 5(i-1) + j - K = 0$$
$$5i + e + j + d + 5i - 5 + j - K = 0$$
$$10i + e + d + 2j - 5 - K = 0$$
$$K = 10i + e + d + 2j - 5$$
But $K$ must be constant for all $i, j$, which is only possible if the coefficients of $i$ and $j$ are zero, i.e., $10i + 2j$ must be constant, which is impossible as $i$ and $j$ vary.

Therefore, it is not possible to make all the numbers in the grid equal by performing such operations.