I other words can I find a number m for any number $n$ such that $\varphi(m)=n$?
It would be great if you could also present a proof or a link to a paper that contains a proof
Edited note: I forgot to mention $n$ must be even as $\varphi(m)$ is always even.
For all $m\geq3$ the number $\varphi(m)$ is even:
For every integer $a$ we have $\gcd(a,m)=\gcd(m-a,m)=1$, and $a\neq m-a$ unless $m=2a$, so $\varphi(m)$ is even if $m$ is odd. If $m=4k+2$ then $$\varphi(m)=\varphi(4k+2)=\varphi(2)\varphi(2k+1)=\varphi(2k+1),$$ which is again odd, and of course if $m=4k$ then $\varphi(m)=\varphi(4k)$ is even.
Also not every even number $n$ is of the form $n=\varphi(m)$, for example $n=14$:
Suppose $\varphi(m)=14$. If $m$ is a prime power, say $m=p^k$, then $$\varphi(m)=\varphi(p^k)=p^{k-1}(p-1).$$ Clearly $k>1$ because $15$ is not prime, so $p=2$ or $p=7$. If $p=2$ then $14=\varphi(m)=2^{k-1}$, a contradiction. If $p=7$ then $14=7^{k-1}\cdot6$, a contradiction. So $m$ is not a prime power; let $a>b>1$ be coprime integers such that $m=ab$. Then $$2\cdot7=14=\varphi(m)=\varphi(a)\varphi(b),$$ so one of $\varphi(a)$ or $\varphi(b)$ is odd; because $b<a$ it follows that $b=2$ and hence $\varphi(b)=1$ and $\varphi(a)=14$ with $a$ odd. Now by the exact same argument we find that $a=cd$ with $c>d>1$ coprime, and hence $d=2$; a contradiction.
No. A natural number which is not in the range of the totient function is called a nontotient. The smallest even nontotient is $14$. See this answer for a proof. (Edit: or Servaes's nice answer!)
