$A, B, C, D, E, F, G, H, I$ are the nine integers from one to nine (not necessarily in order). They satisfy the following constraints:

$A + B +C +D = 20$
$B+C+D+E+F = 20$
$D+E+F+G+H = 20$
$F+G+H+I=20$

What values are taken by each of `A` to `I`?

Since $A, B, C, D$ and $F, G, H, I$ sum to 40, $E$ must be 5.
If `E` is 5, subtracting equation 1 with equation 2, `A - 5 - F = 0`; subtracting equation 3 with equation 4, `D +5 - I = 0`. So, any combinations that satified these two equations can be the answer. For example, `A=9`, `F=4`, `D=2`, `I=7`.
With this, only `B`, `C`, `G`, `H` are left. Subtracting equation 2 with equation 3, `B + C = G + H`. There are again many combinations depending on what you chose for `A`, `F`, `D`, `I`. If we used the above numbers, what’s left is `1, 3, 6, 8`. So, one answer can be `B=1`, `C=8`, `G=6`, `H=3`.