Congratulations to Wei Zhang, age 18 from Merchiston Castle School for the superb solution given below. Andrei Lazanu, age 14, School No. 205, Bucharest, Romania submitted a similar solution.

Suppose there is a fraction 1/n , where n belongs to {1,2,3,4,5,6}.
1/1: 1x
=1 so x=1
1/2: 2x
=1 so x=4
1/3: 3x
=1 so x=5
1/4: 4x
=1 so x=2
1/5: 5x
=1 so x=3
1/6: 6x
=1 so x=6
As I showed above, 1/n is in the set {1,2,3,4,5,6}.

Now we have a/n = a ×1/n , where a belongs to {0,1,2,3,4,5,6}, this is a number from the set {0,1,2,3,4,5,6} times a number from the set of {1,2,3,4,5,6}. Therefore we will get a set of numbers for a/n , which is {0,1,2,3,4,5,6}. Solving the equation
 1
x
 +  1
y
 =  1
x+y
is equivalent to solving x2+xy+y2=0.
when x=1 y2+y+1=0 so y2+y=6 y=2 or y=4
when x=2 y2+2y+4=0 so y2+2y=3 y=1 or y=4
when x=3 y2+3y+2=0 so y2+3y=5 y=6 or y=5
when x=4 y2+4y+2=0 so y2+4y=5 y=1 or y=2
when x=5 y2+5y+4=0 so y2+5y=3 y=3 or y=6
when x=6 y2+6y+1=0 so y2+6y=6 y=3 or y=5.

We have found the six solution pairs (1,2), (1,4), (2,4), (3,6), (3,5), (5,6). The equation is symmetric in x and y so there are six corresponding solutions when we exchange x and y. We don't allow negative solutions (e.g. x=1, y=−3) because we are working entirely in the set {0,1,2,3,4,5,6}. If we work with real numbers, we think of solving the quadratic equation x2+yx+y2=0 as an equation in x. The formula for the solution of this quadratic equation involves y2−4y2=−3y2 < 0 which has no real values, therefore there is no real solutions for x.