<s>[INST] Any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$. My solution Let $n\in\mathbb{Z}^{+}$, then there exists $k\in\mathbb{Z}_0^+$, such that $n=5k + i, i\in\{0,1,2,3,4\}$. Now analyzing by cases we have:

*

*If $i=0$, then
\begin{align*}
        n = 5k \Rightarrow n = 5k + 4(0).
\end{align*}

*If $ i = 1 $, then
\begin{align*}
       n & = 5k + 1 \\
         & = 5k-5(3) +5(3) +1 \\
         & = 5(k-3) + 15 + 1 \\
         & = 5(k-3) +16 \Rightarrow n = 5(k-3) +4(4).
\end{align*}

*If $ i = 2 $, then
\begin{align*}
   n & = 5k + 2 \\
     & = 5k-5(2) +5(2) +2 \\
     & = 5(k-2) + 10 + 2 \\
     & = 5(k-2) +12 \Rightarrow n = 5(k-2) +4(3).
\end{align*}

*If $i=3$, then
\begin{align*}
   n & = 5k + 3 \\
     & = 5k-5 + 5 + 3 \\
     & = 5(k-1) +8 \Rightarrow n = 5(k-1) +4(2).
\end{align*}

*If $i=4$, then
\begin{align*}
    n = 5k + 4 \Rightarrow n = 5k + 4(1).
\end{align*}
Thus, every positive number can be expressed as a linear combination of $5$ and $4$. Now using that $n>11$, so we have:
\begin{align*}
n       &> 11 \\
5k + i  &> 5(2) +1 \\
5k-5(2) &> 1-i \\
5 (k-2) &> 1-i \\
k-2     &> \frac{1-i}{5} \\
k       &> 2+\frac{1-i}{5}.
\end{align*}
So by increasing over the values ​​that $ i $ takes, we have:
\begin{align*}
k &> 2+ \frac{1-i}{5} \geq 2+ \frac{1-0}{5}\\
k &> 2 + 0.2 = 2.2
\end{align*}
But $k\in\mathbb{Z}_0^+ \Rightarrow k \geq 3 \Rightarrow n \geq 15 $. Thus we have that every positive integer greater than or equal to $15$ is a non-negative linear combination of $5$ and $4$.
Finally, let's look at the cases that are still unverified, which are $12$, $13$ and $14$.
\begin{align*}
12 &= 5(0) +4(3) \\
13 &= 5(1) +4(2) \\
14 &= 5(2) +4(1).
\end{align*}
Therefore, any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
 [/INST] You also can prove it by induction on $n>11$.
Here is how the inductive step goes:
Suppose that for some $n>11$, one has $\;n=5x+4y$, $x,y \ge 0$.

*

*If $y >0$, as $1=5-4$, you can write $n+1=5(x+1)+4(y-1)$, and $y_1\ge 0 $;

*If $y=0$, $n=5x$, and since $n>11$, you have $x\ge 3$. Now you also can write $1=16-15$, so
$$n+1=5(x-3)+4\cdot 4\qquad(\text{note that } x-3\ge 0)$$
</s>