Some variant of the following result is proved in most books in elementary number theory.
Theorem: Let $a$ and $b$ be relatively prime positive integers. If $c >ab$, then there exist positive integers $x$ and $y$ such that $ax+by=c$.
The proof is not difficult. It is not quite a one-liner, largely because I am not a man of few words.
There are integers $x_0,y_0$, such that $ax_0+by_0=c$. Consequently, all integer solutions of the equation $ax+by=c$ have the shape $x=x_0+bt$, $y=y_0-at$, where $t$ ranges over the integers. To produce a positive solution, we want to find $t$ such that $x>0$ and $y>0$.
So we need $y_0-at >0$, $x_0+bt>0$, or equivalently
$$-\frac{x_0}{b}<t <\frac{y_0}{a}.$$
The interval $(-x_0/b,y_0/a)$ has width $y_0/a+x_0/b$. which simplifies to $(ax_0+by_0)/ab$, that is, $c/ab$. If $c/ab>1$, then the interval is guaranteed to contain an integer $t$, and we are finished.
If $a$ and $b$ are not relatively prime, let $d=\gcd(a,b)$, and let $a=da'$, $b=db'$. Let $c$ be a multiple of $d$, say $c=c'd$. Then if $c'>a'b'$, $c$ is representable as a positive linear combination of $a$ and $b$. Equivalently, every multiple $c$ of $d$ which is greater than $ab/d$ is a positive linear combination of $a$ and $b$.
In particular, the $\text{lcm}(a,b)$ term in your expression puts us past the numbers that do not have a positive representation. The bound $ab/d$ is quite sharp.
The generalization to "many periods" is straightforward. Getting sharp bounds on the smallest number $m$ such that all $c$ that satisfy the necessary divisibility condition are representable may not be easy.