My attempt: I have an idea about frobenius numbers but I didn't want to use the result.So i started with the basics: let two numbers be a, b such that gcd(a,b)=1.
Now $a>=1$and $ab>=b$.since $b.a+a.0=ab$so from here I can get an intuition that numbers after$ ab$can be written as a linear combination.
Now I try to prove that $ab+k=ax+by$where $x, y, k$are all positive numbers. Now we note one idea that gcd(a,b)=1 and so $au+bv=1 $(by bezouts identity) . And so $k=am+bn$ Where m and n can be positive or negative. Now suppose m is negative and n is positive. Then $ab+k=a(b+m)+bn$if the coefficient of a is still negative then we can have $a(b+m+yb)+b(n-ya)$.we chose $b+m+yb<b.y<-\frac{m}{b}$.now $-ya>\frac{ma}{b}$. And $n-ya>\frac{k}{b}$which is positive.
$1)$if there are flaws in my answer can someone find it out and also how do I do the same proof using Euclids division algorithm
$2)$can someone drop some hints about how do I do it for n numbers
Edit1:suppose there are $n$ Coprime numbers. $a_q,a_2,a_3,..a_n$.Now I do the same thing $a_1(a_2+..a_n)>=a_1$ and similarly $a_1(a_2+..a_n)+a_2.0+a_3.0+..a_n.0=a_1(a_2+..a_n)$ .
So I try to prove that a number greater than this will be represented. They are coprime so their Gcd is 1 .hence $a_1.x_1+a_2.x_2+..a_n.x_n =1$.$a_1.y_1+..a_n.y_n=k$.(k has the same definition as above.)
Now, consider the worst case that$ y_2,..y_n$ are negative . By similar arguments in case (1) i show that the coefficients are all positives.