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There is a new upper bound of 254/67 (= 3.79104477...).

Define 6 sets of cardinality 4:

X1={-B+A, 0, A, B}
Y1={0, A, B-A, B}

X2={-B, -B+3A, B-2A, B+A}
Y2={-2B+A, -A, B+A, 3B-A}

X3={-6B+2A, -2B-2A, 2B+3A, 6B-A}
Y3={-2B-2A, -B+6A, 2B-6A, 3B+2A}

then we already know that in the in the grid X1 x Y1 if the sum in the central AxB is 4+$\epsilon$ $4+\epsilon$ the sum in the surrounding (2B-A)x(B-2A) is 4+3$\epsilon$,$\geq4+3\epsilon$,

similarly in the in the grid (X1 $\cup$ X2) x (Y1 $\cup$ Y2) if the sum in the central AxB is 19/5+$\epsilon$ $19/5+\epsilon$ then the sum in the surrounding (2B-5A)x(5B-2A) is -19/5-21$\epsilon$,$\leq-19/5-21\epsilon$,

last, in the in the grid (X1 $\cup$ X2 $\cup$ X3) x (Y1 $\cup$ Y2 $\cup$ Y3) if the sum in the central AxB is 254/67+$\epsilon$ $254/67+\epsilon$ then the sum in the surrounding (12B-3A)x(3B-12A) is 254/67+135$\epsilon$.$\geq254/67+135\epsilon$.

All of the above claims are easily verifiable with the tools already described in the previous answers and comments. I wonder if one can find sets X4 and Y4 (with 4 elements each?) to further improve the bound and maybe spot a general pattern.

There is a new upper bound of 254/67 (= 3.79104477...).

Define 6 sets of cardinality 4:

X1={-B+A, 0, A, B}
Y1={0, A, B-A, B}

X2={-B, -B+3A, B-2A, B+A}
Y2={-2B+A, -A, B+A, 3B-A}

X3={-6B+2A, -2B-2A, 2B+3A, 6B-A}
Y3={-2B-2A, -B+6A, 2B-6A, 3B+2A}

then we already know that in the in the grid X1 x Y1 if the sum in the central AxB is 4+$\epsilon$ the sum in the surrounding (2B-A)x(B-2A) is 4+3$\epsilon$,

similarly in the in the grid (X1 $\cup$ X2) x (Y1 $\cup$ Y2) if the sum in the central AxB is 19/5+$\epsilon$ then the sum in the surrounding (2B-5A)x(5B-2A) is -19/5-21$\epsilon$,

last, in the in the grid (X1 $\cup$ X2 $\cup$ X3) x (Y1 $\cup$ Y2 $\cup$ Y3) if the sum in the central AxB is 254/67+$\epsilon$ then the sum in the surrounding (12B-3A)x(3B-12A) is 254/67+135$\epsilon$.

All of the above claims are easily verifiable with the tools already described in the previous answers and comments. I wonder if one can find sets X4 and Y4 (with 4 elements each?) to further improve the bound and maybe spot a general pattern.