, 2024

Solution:

Suppose the initial grid is of the format shown below:
$$\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]$$
After the transformation, we end with
$$\left[\begin{array}{lll} a_{n} & b_{n} & c_{n} \\ d_{n} & e_{n} & f_{n} \\ g_{n} & h_{n} & i_{n} \end{array}\right]=\left[\begin{array}{ccc} b+d & a+c+e & b+f \\ a+e+g & b+d+f+h & c+e+i \\ d+h & g+e+i & f+h \end{array}\right]$$
Since $d \neq f$, $a_{n}=b+d \neq b+f=c_{n}$. By symmetry, no two corners on the same side of the grid may both be $23$ after the transformation.
Since $c \neq g$, $b_{n}=a+c+e \neq a+e+g=d_{n}$. By symmetry, no two central-edge squares sharing a corner may both be $23$ after the transformation.
Assume for the sake of contradiction that $e_{n}=23$. Because $a_{n}, c_{n}, g_{n}, i_{n}<e_{n}$, none of $a_{n}, c_{n}, g_{n}, i_{n}$ can be equal to $23$. Thus, $3$ of $b_{n}, d_{n}, f_{n}, h_{n}$ must be $23$. WLOG assume $b_{n}=d_{n}=f_{n}=23$. This is a contradiction however, as $b_{n} \neq d_{n}$. Thus, $e_{n} \neq 23$.
This leaves the case with two corners diametrically opposite and two central edge squares diametrically opposite being $23$. WLOG assume $a_{n}=b_{n}=h_{n}=i_{n}=23$.
Thus, $92=4 \cdot 23=a_{n}+b_{n}+h_{n}+i_{n}=(b+d)+(a+c+e)+(e+g+i)+(f+h)=(a+b+c+d+e+f+g+h+i)+e$. Since $a+b+c+d+e+f+g+h+i=74$, this means that $e=92-74=18$.
One possible example of $18$ working is $\left[\begin{array}{lll}4 & 16 & 2 \\ 6 & 18 & 7 \\ 1 & 17 & 3\end{array}\right]$. Thus the only possible value for the center is $18$.