An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$.
We will prove that $k_{\max }=45$.
We number the columns and the rows and we select all possible $3^{2}=9$ choices of an odd column with an odd row.

Collecting all such pairs of an odd column with an odd row, we double count some squares. Indeed, we take some $3^{2}$ squares 5 times, some 12 squares 3 times and there are some 4 squares (namely all the intersections of an even column with an even row) that we don't take in such pairs.

It follows that the maximal total sum over all $3^{2}$ choices of an odd column with an odd row is

$$
5 \times(17+18+\cdots+25)+3 \times(5+6+\cdots+16)=1323
$$

So, by an averaging argument, there exists a pair of an odd column with an odd row with sum at most $\frac{1323}{9}=147$.

Then all the other squares of the array will have sum at least

$$
(1+2+\cdots+25)-147=178
$$

But for these squares there is a tiling with $2 \times 2$ arrays, which are 4 in total. So there is an $2 \times 2$ array, whose numbers have a sum at least $\frac{178}{4}>44$. So, there is a $2 \times 2$ array whose numbers have a sum at least 45 . This argument gives that

$$
k_{\max } \geq 45 .
$$

We are going now to give an example of an array, in which 45 is the best possible. We fill the rows of the array as follows:

| 25 | 5 | 24 | 6 | 23 |
| :---: | :---: | :---: | :---: | :---: |
| 11 | 4 | 12 | 3 | 13 |
| 22 | 7 | 21 | 8 | 20 |
| 14 | 2 | 15 | 1 | 16 |
| 19 | 9 | 18 | 10 | 17 |

We are going now to even rows:
In the above array, every $2 \times 2$ subarray has a sum, which is less or equal to 45 . This gives that

$$
k_{\max } \leq 45 .
$$

A combination of (1) and (2) gives that $k_{\max }=45$.

[^6]C3. Anna and Bob play a game on the set of all points of the form $(m, n)$ where $m, n$ are integers with $|m|,|n| \leqslant 2019$. Let us call the lines $x= \pm 2019$ and $y= \pm 2019$ the boundary lines of the game. The points of these lines are called the boundary points. The neighbors of point $(m, n)$ are the points $(m+1, n),(m-1, n),(m, n+1),(m, n-1)$.

Anna starts with a token at the origin (0,0). With Bob playing first, they alternately perform the following steps: At his turn, Bob deletes two points on each boundary line. On her turn Anna makes a sequence of three moves of the token, where a move of the token consists of picking up the token from its current position and placing it in one of its neighbors.

To win the game Anna must place her token on a boundary point before it is deleted by Bob. Does Anna have a winning strategy?
[Note: At every turn except perhaps her last, Anna must make exactly three moves.]