Arrange the numbers $1$ to $9$ to replace letters $A$ to $I$ so:
$(A+B+C+D)-(E+F+G+H) = I$
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We know there are 4 evens and 5 odds in the set 1:9.
Let's assume I is odd, then the left side of the equation consists of 4 odds and 4 evens. We can rearrange the equation to separately sum/substract the 4 odds together, which will be even, and 4 evens together, which will be even. Summing those together implies the left hand side of the equation will be even, a contradiction.
Let's assume I is even, then the left side of the equation consists of 5 odds and 3 evens. We can rearrange the equation to separately sum/subtract the 5 odds together, which will be odd, and 3 evens, which will be even. Summing those two together implies the left hand side of the equation will be odd, a contradiction
There is no solution to the provided equation!
Thought I'd share my algebraic solution
(A + B + C + D) - (E + F + G + H) = I
A + B + C + D + E + F + G + H + I = 45
45 - (A + B + C + D + E + F + G + H) = I
45 - (A + B + C + D + E + F + G + H) = (A + B + C + D) - (E + F + G + H)
45 - 2(A + B + C + D) = 0
A + B + C + D = 22.5
Meaning A, B, C, D are not all integers, so there is no solution
Adding or subtracting 2 odd/even numbers is even. From 1 - 9 we have 5 odd numbers and 4 even numbers. The above equation
A + B + C + D - (E + F + G + H) = I
Can be rearranged as
A + B + C + D - (E + F + G + H + I) = 0
Adding or subtracting 4 even numbers will result in even number similarly adding or subtracting 4 odd numbers will be even. So we are left with 1 odd number
4 ( odd ) +/- 4 ( even ) +/- 1 odd = 0
Even +/- Even +/- odd = 0
Which is not possible since even plus odd is always odd.