We can use an unlimited supply of 4-cent and 7-cent postage stamps to make (exactly) any amount of postage that is 18 cents or more. $P(n)$: postage of exactly n cents can be made using only 4 cents and 7 cents
Basis ($n=18$)
$1 \cdot 4 + 2\cdot7 = 18$, thus $P(18)$ holds
Induction step:
Let $i$ be an arbitrary number such that $i \geq18$ and suppose that $k, L \in \mathbb N$ such that $i = 4\cdot k + 7 \cdot L$ k representing the # of 4 stamps and L representing the # of 7 stamp
I was told I have to prove 2 cases with them being $L > 0$ and $L =0$. I don't get how they got the two cases. I'm only aware that I have to prove that any number $i \geq 18$ can be achieved with 4 stamps and 7 stamps. 
Could someone explain this proof?
 A: Remember that in general, the structure of the inductive step would be like this:

Let $i\in\mathbb N$ and $i \geq 18,$ and suppose $P(i)$ is true. We wish to show that $P(i+1)$ is true.
Since $P(i),$ ...
...
...   and therefore $P(i+1).$ This proves the inductive step.

Your problem is to fill in the "..." parts with a valid mathematical argument.
Your instructor's hint about cases $L > 0$ and $L=0$ gives you a little more detail about how the inductive step can be structured.
The hint says to try this:

Let $i\in\mathbb N$ and $i \geq 18,$ and suppose $P(i)$ is true. We wish to show that $P(i+1)$ is true.
Since $P(i),$ there exist $k, L \in \mathbb N$ such that $i = 4\cdot k + 7 \cdot L.$ Consider the following two cases, which cover all possible values of $L$:
Case 1: Suppose $L>0.$ Then ...
...
... and therefore $P(i+1).$
Case 2: Suppose $L=0.$ Then ...
...
... and therefore $P(i+1).$
Hence in either case we have shown $P(i+1).$ This proves the inductive step.

You still have to fill in the "..." parts. With two separate arguments.
But the only way to learn how to do this is to start trying.
By the way, the details of your question seem to imply that your book and/or instructor consider the least element of $\mathbb N$ to be $0.$  In some other contexts, people consider the least element of $\mathbb N$ to be $1$ (and therefore $0 \not\in\mathbb N$). I think what you have written so far is probably perfectly OK, but in general, when dealing with $\mathbb N$ make sure you know which way it has been defined.
A: INDUCTION STEP:
Suppose you can make $N$ by $4m + 7n$.
Then you can make $N+1$ by doing $4(m+2) + 7(n-1) = 4m +8 + 7n -7 = 4m + 7n + 1 = N + 1$.
That'd a good induction step.... except what if $n < 1$?
Well then you could do $N = 4m + 7n = 4(m -5) + 7(n+3)=4m -20 + 7n + 21 = 4m + 7n + 1$.
That'd a good alternative induction step... except what if $m < 5$?   
Well,... then you'd do the first one.  Any any case you can either do the first induction step, or the second induction step, or take your choice of which one to use.
Okay, what if both $m < 5$ and $n < 1$. Well, ... then we'd be screwed.  
But  If that were the case $m \le 4;n=0$ and $N \le 4*4 + 7*0 = 16$.
So, so long as $N > 16$ we will have our choice of induction steps and we can always get $N+1$.  
... Provided our base case is $N > 16$.
BASE CASE
If $N=17$ we .... can't do it. 
If $N = 18$ we can do it with $m = 1$ and $n= 2$ and so $N = 4m + 7n = 4*1 + 7*2 = 4+14 = 18$.
Hence we are done by induction.
[Note: We can do $16 = 4*4 + 7*0$ but to get $17$ we have to do $4*(4-5) + 7*(0+3)= 4(-1) + 7*3$.  And we can also do $14$ to $15$ to $16$ by $14 = 4*0 + 7*2; 15 = 4*(0+2) + 7*(2-1) = 4*2 + 7*1; 16 = 4*(2+2)+7*(1-1) = 4*4 + 7*0$.  But then we hit a wall because we have both $m < 5$ and $n < 1$ so we can't do the induction step.]
A: \begin{eqnarray*}
18=1 \times 4 + 2 \times 7 \\
19=3 \times 4 + 1 \times 7 \\
20=5 \times 4 + 0 \times 7 \\
21=0 \times 4 + 3 \times 7 \\
\end{eqnarray*}
Now add $4$ to these.
A: For the record, why don't we prove that if $a$, $b$ are relatively prime positive integers then every integer $> ab$ is of the form $ u a + v b$, with $u$,$v$  integers  $>0$. Indeed, take $k> ab$ integer. Then (exactly) one of the integers $a$, $2 a$, $\ldots$, $b a$ will be $\equiv k \mod b$, let that be $u a$. So $k - u a = v b$, with $v > 0$. 
