Assume exist $36$ distinct numbers with the requested property. Denote $a_{ij}$ the number situated at the intersection of the row $i$ and the column $j;\;i,j\in\{1,2,3,4,5,6\}$. Using the given conditions in the first row, we obtain: $\begin{cases}a_{11}+a_{12}+a_{13}+a_{14}+a_{15}\in\{2022,2023\}\\a_{12}+a_{13}+a_{14}+a_{15}+a_{16}\in\{2022,2023\}\\a_{11}\ne a_{16}\end{cases}\Longrightarrow a_{11}- a_{16}\in\{-1,+1\}\quad(1)$. Applying the same method in the last row, respectively in the first and the last column, we obtain: $a_{61}-a_{66}\in\{-1,+1\};\;a_{11}-a_{61}\in\{-1,+1\};\;a_{16}-a_{66}\in\{-1,+1\}\quad(2)$. WLOG, we can consider $\min\{a_{11},a_{16},a_{61},a_{66}\}=a_{11}$. From (1) and (2) results: $\begin{cases}a_{16}=a_{11}+1\\a_{61}=a_{11}+1\end{cases}\Longrightarrow a_{16}=a_{61}$, contradiction with the initial condition (the $36$ numbers are distinct). Hence: it is not possible to arrange $36$ distinct numbers with the requested property.

No. Proof: BWOC, assume that such grid exists. We let it's corner be $a,b,c,d$ respectively. Notice that $a \neq b \neq c \neq d$, then $|a-b| = |b-c| = |c-d| = |d-a| = 1$. WLOG assume that $b=a+1, d=a-1$. Then by $|b-c| = |c-d| \implies b^2 - 2cb + c^2 = c^2 + d^2 - 2cd \implies b^2 - d^2 = 2c(b-d) \implies 4a = 2c(2a) \implies c=a$, contradiction. QED.