The Coffee House

Every row/columnn contains the numbers 1 to 9 exactly once. This means that every row and column must add up to 45.
In particular: the top row, the bottom row, the left-most column, and the right-most column, must sum to 4x5 = 180.
The numbers around the perimeter sum to (28 + 12 + 22 + 14 + 23 + 17 + 20 + 10) = 146.
This double-counts the four corners; therefore, the four corners must add up to 34.
The only way in which this can work is if the four corners are two 8s and two 9s.
Therefore, one opposing pair of corners has two 8s, and the other has two 9s

The middle column adds up to 45.
The middle column (minus the two ends) adds up to (10 + 13 + 10) = 33.
Therefore, the two ends of the middle column must add to (45 - 33) = 12.
There are three combinations that make this work: 5 and 7; 4 and 8; and 3 and 9.
However, the last two are impossible, as the 8 and 9 are already used in the corners.
Therefore, the middle top and middle bottom must be a 5 and a 7

If the middle bottom was a 5, then the two cells either side of it would add up to 12.
However, they can't be a 5 and 7, because the 5 is already used in the middle.
They can't be a 4 and 8, because the 8 is already in one corner - and they can't be a 3 and 9, because the 9 is already used in the other corner.
Therefore, there's no combination that makes this work.
This means the middle bottom must be a 7, and the top middle must be a 5

The two cells either side of the top middle add up to 7.
They can't be a 2 and a 5, because the 5 is already used in the middle.
Therefore, the two cells either side of the 5 at the top must be either a 1 and a 6, or a 3 and a 4

The two cells either side of the bottom middle add up to 10.
They can't be a 3 and 7, because the 7 is already used in the middle.
They can't be a 2 and 8, because the 8 is already in one corner - and they can't be a 1 and 9, because the 9 is already used in the other corner.
This just leaves 4 and 6.
Therefore, the two cells either side of the 7 at the bottom must be a 4 and 6

The middle row contains two pairs which add up to 10.
They can't be a 3 and a 7, because the 7 is already used at the bottom.
Therefore, the 3 must be somewhere in the middle bit, which adds up to 13

Now, the remaining numbers in the middle row consist of three pairs, which add up to 10.
These will be: a 1 and a 9; a 2 and an 8; and a 4 and a 6.
The 4 and the 6 cannot be at the top, because there will be either a 4 or a 6 to the side of the 5.
The 4 and the 6 cannot be at the bottom, because there's both a 4 and a 6 to the side of the 7.
Therefore, the 4 and the 6 are in the middle, together with the 3

The four rows on the right add up to (4 x 45) = 180.
These four rows (minus the two cells to the right of the 5 and 7) add up to (22 + 10 + 13 + 23 + 14 + 31 + 19 + 23 + 15) = 170.
Therefore, the cell to the right of the 5, and the the cell to the right of the 7, must add up to 10.
This is only possible if the cell to the right of the 5 is either a 4 or a 6.
Therefore, the number to the right of the 5 is either a 4 or a 6, and the number to the left of the 5 is either a 1 or a 3