Olimpiade Italiana di Matematica, Cesenatico

Solution:

For every admissible $k$, the minimum number of paths needed to determine $a_{k}$ with certainty is two.

One path is not enough
Consider a path, and denote by $c_{i}$ the number of cells containing the value $a_{i}$ traversed by the path. We observe that $c_{i} \geq 1$ for every $i$ and the sum equals
$$S = c_{1} a_{1} + \ldots + c_{k} a_{k} + \ldots + c_{2022} a_{2022}.$$
Now Barbara knows the coefficients $c_{1}, \ldots, c_{2022}$, and would like to determine the number $a_{k}$ uniquely. Suppose we have found this value of $a_{k}$, and choose an index $h \neq k$. One verifies that the sum $S$ does not change if $a_{k}$ is replaced by $a_{k} + c_{h}$, and $a_{h}$ is replaced by $a_{h} - c_{k}$, leaving all the other $a_{i}$ unchanged. This shows that there is never a unique value of $a_{k}$ that is compatible with the given sum.

Two paths are enough when $2 \leq k \leq 2022$
Consider the path $P$ that traverses the first row to the right until it reaches $a_{k}$, then moves down into the second row and goes back until the first column, then moves down to the last row and finally moves to the right until the destination. Also consider the path $P'$ analogous to the previous one, with the only difference that the first "turn" happens one move earlier, that is, when the path reaches the cell with $a_{k-1}$ (in the case $k=2$ this means that the path $P'$ moves directly downward).

The following figure represents the two paths in the special case $k=5$ (for simplicity we have written $a_{n}$ instead of $a_{2022}$).

Path $P$

Path $P'$

The only difference between the two paths is that $P$ has two more cells with $a_{k}$. Denoting by $S$ and $S'$ the sums over the two paths, it follows that
$$S - S' = 2 a_{k}$$
from which the value of $a_{k}$ is immediately obtained.

Two paths are enough when $k=1$
Consider the two paths $P$ and $P'$ that start by going from the initial cell to cell $(4,4)$ in this way.
- Path $P$ makes in succession two moves to the right, one downward, one to the right, and finally two downward.
- Path $P'$ traverses the entire first row, then descends into the second row and goes back to the second cell of the second row, then makes one move downward, one to the left, one downward, and finally traverses the fourth row to the right up to cell $(4,4)$.

From cell $(4,4)$ to the end the two paths continue in the same way, always alternating one move to the right and one downward. In this way the two sums turn out to be
$$S = a_{1} + a_{2} + 2 a_{3} + 3 a_{4} + 2(a_{5} + \ldots + a_{2022})$$
and
$$S' = a_{1} + 2 a_{2} + 4 a_{3} + 6 a_{4} + 4(a_{5} + \ldots + a_{2022})$$
from which it follows that
$$2S - S' = a_{1}$$

Path $P$

Path $P'$