# there is an integer $N_0$ such that for all $N\geq N_0$ the equation $ax+by=N$ can be solved with both $x$ and $y$ nonnegative integers.

Let $a$ and $b$ be two relatively prime positive integers. Prove that every sufficiently large positive integer $N$ can be written as a linear combination $ax+by$ of $a$ and $b$ where $x$ and $y$ are both nonnegative, i.e. there is an integer $N_0$ such that for all $N\geq N_0$ the equation $ax+by=N$ can be solved with both $x$ and $y$ nonnegative integers.

Proof: Let $a,b$ be positive integers. Assume that $a$ and $b$ are relatively prime. i.e. $\gcd(a,b)=1$. i.e. $ax+by=1$ for some $x,y\in \mathbb{Z}$. How can I restrict this result to $x,y$ are positive integers?

Let $N$ be an integer such that $N\geq N_0$. Then $$N = N(ax+by) = a(Nx)+b(Ny)$$ So $n$ can be written as a linear combination of $a$ and $b$.

Hint: Consider $N_0 = ab-a-b+1.$ To prove this consider the numbers $0,b,2b,...,(a-1)b$ and use the fact that they are all distinct modulo $a$ and thus represent all possible remainders modulo $a.$
We only need to show that for all integer $$N\ge ab+1$$, the equation $$ax+by=N$$ can be solved with both $$x$$ and $$y$$ positive integers. Note that the difference of the statement here with that in the question is that we need $$x$$ and $$y$$ to be positive.
As @dezdichado mentioned, the $$a$$ numbers $$b,2b,...,ab$$ are all distinct modulo $$a$$ since the difference of any pair of them cannot be divided by $$a$$, and thus they represent all possible remainders modulo $$a$$.
Now for any integer $$N\ge ab+1$$, the remainder of $$N$$ molulo $$a$$ must be the same as that of $$by$$ molulo $$a$$, with $$1\le y\le a$$ an integer. That is, $$a\mid(N-by)$$. Then there exists a nonnegative integer $$x$$ such that $$N=ax+by$$, and it is easy to verify that $$x$$ cannot be zero. We are done.