2x+y+3z=59

y = 59-2x-3z

y = 59-(2x+3z)

y will take on its maximum possible value when the 
least amount is subtracted from it.

2x+3z is what is subtracted from it.  Since all the
letters represent positive integers, 2x+3z is least
when x=z=1, which is when 

y = 59-(2x+3z) = 59-(2*1+3*1) = 59-(2+3) = 59-5 = 54

So the maximium positive integer y can be is 54.

----------

Let's see in y can be the minimum possible positive integer,
which is 1.

2x+y+3z=59

We substitute y=1 to see if this is possible.

2x+1+3z=59

2x+3z=58

2x = 58-3z

x = (58-3z)/2

We are told that x must be a positive integer.
x will be a positive integer if and only if 3z is even.
If we choose z as any even integer from 2 up through 18,
3z will be even and since 58 is even, 58-3z will be even

So for instance, if we choose z = 2 then
x = (58-3z)/2 = (58-3*2)/2 = (58-6)/2 = 52/2 = 26

So y will take on its minimum value of 1 when x=26 and z=2

[Incidentally y will take on its minimum value of 1 in
any of these cases:

z=2 and x=26
z=4 and x=23
z=6 and x=20
z=8 and x=17
z=10 and x=14
z=12 and x=11
z=14 and x=8
z=16 and x=5
z=18 and x=2

Edwin