Non-biodegradable

Suppose that $N \in \mathbb{N}$ with $(N,2033) = 1$ . Note that $2033 = 19\times107$ . Since $\mathbb{Z}/19\mathbb{Z}$ is a field, we can find $1 \le j \le 18$ such that $Nj \, \equiv\, 12 \, \equiv\, 107$ modulo $19$ .

1. If $N \ge 108$ then $Nj - 107 \in \mathbb{N}$ is congruent to $0$ modulo $19$ , and hence we can write $Nj = 107 + 19c$ for some $c \in \mathbb{N}$ . Thus $\frac{N}{2033} \; = \; \frac{1}{19j} + \frac{c}{107j}$ and (since $1 \le j \le 18$ ) we deduce that $1 \,\le\, 19j\,,\,107j\,<\, 2033$ .

2. If $N \ge 54$ and $N \not\equiv 12$ modulo $19$ , then we must have $j \ge 2$ . But then $Nj - 107 \in \mathbb{N}$ is congruent to $0$ modulo $19$ , and hence we can write $Nj = 107 + 19c$ for some $c \in \mathbb{N}$ . Thus $\frac{N}{2033} \; = \; \frac{1}{19j} + \frac{c}{107j}$ and, again, $1 \,\le\, 19j\,,\,107j\,<\, 2033$ .

The numbers $54 \le N \le 107$ which are congruent to $12$ modulo $19$ are $69,88,107$ , and $107$ is not coprime to $2033$ . Thus we know that, for any integer $N \ge 89$ which is coprime to $2033$ , $\frac{N}{2033}$ can be written in the desired form. It remains to confirm that $88$ cannot be dealt with similarly.

Suppose that we can write $\frac{88}{2033} \,=\, \frac{a}{b} + \frac{c}{d}$ where $a,b,c,d \in \mathbb{N}$ , $(a,b) = (c,d) = 1$ and $1 \le b,d < 2033$ . Then $88bd = 2033(ad + bc)$ , and hence both $19$ and $107$ divide $88bd$ , and hence both divide $bd$ . Since $b,d < 2033$ , we deduce that $19$ must divide one of $b,d$ , with $107$ dividing the other. Without loss of generality, assume that $19$ divides $b$ , and that $107$ divides $d$ , so that $b = 19u$ and $d = 107v$ where $u,v \in \mathbb{N}$ are such that $(a,19u) = (c,107v) = 1$ and $1 \le u < 107$ and $1 \le v < 19$ . Then $88uv = 107av + 19bu$ . From this we deduce that $u | 107av$ . Now $1 \le u < 107$ , so $(u,107) = 1$ . Moreover $(u,a)= 1$ . Thus we deduce that $u|v$ . Similarly $v | 19bu$ . Now $1 \le v < 19$ , so $(v,19) = 1$ . Moreover $(v,b) = 1$ . Thus we deduce that $v | u$ . Hence it follows that $u=v$ , so that $88u = 107a + 19b$ where $1 \le u < 19$ . Then $12u \; \equiv \; 88u \; \equiv \; 107a \; \equiv \; 12a \qquad \qquad (19)$ so we deduce that $u \equiv a$ modulo $19$ . From this it is clear that $a \ge u$ , and hence that $107a + 19b \; > \; 107a \; > \; 88u$ which is impossible. Thus we deduce that $\frac{88}{2033}$ cannot be expressed in this way, and hence the answer to the question is $88$ .

A computer search showed me the need for one of the two fractions to be of the form $\frac{1}{19j}$ ; once I had that to aim for, the rest fell into place.

So first I restated the problem: $Nz=19x+107y,1<=z<=18,x>0,y>0$ To change the question into this, first set $\frac{a}{b}+\frac{c}{d}=\frac{N}{2033}$ , where $(a,b)$ and $(c,d)$ are each coprime. Now I will prove that b*d is a multiple of 2033. If it is not, then by adding the two fractions together with b*d as denominator and setting it equal to $N/2033$ , N must not be coprime to 2033. Now, using the coprimality of a and b, you can show that b is a factor of d. Using the coprimality of c and d, you can show that d is a factor of b. That proves b=d, which then will reduce to my restatement above, where x=a,y=c,and z=b and d. N cannot be more than 113 because then setting z=18 $zN$ is more than 2033, which leads to there being at least 19 options for y, at least one of which must lead to a solution.Then trial and error lead me to 88. I know this isn't so clear, but it's how I did it.