Is there an integer $N>0$ such that $\varphi(n) = N$ has infinitely many solutions? Let $\varphi: \mathbb{N} \to \mathbb{N}$ be the totient function. 
Is there an integer $N > 0$ such that there are infinitely many integers $n > 0$ such that
$$\varphi(n) = N?$$

No. 
We use the fact that if $n = p_1^{e_1} \cdots p_k^{e_k}$, with the $p_i$ different primes and the $e_i$ positive, we have $\phi(n) = p_1^{e_1-1}(p_1-1) \cdots p_k^{e_k-1}(p_k-1)$.
Now consider an integer $N>0$ and suppose that $\phi(n) = N$. Then $n$ cannot be divisible by primes larger than $N+1$, since if $q > N+1$ divides $n$, we have $N+1 \leq q-1 \mid \phi(n) = N$, contradiction. The set of primes that may divide $n$, therefore, is finite. Let $p$ be such a prime. Then if $p^k \mid n$ we have $p^{k-1} \mid N$, which leaves a finite number of possibilities for $k$. Thus the number of primes that may divide $n$ is bounded and the exponent of each prime is bounded as well, thus there are at most finitely many $n$ for which $\phi(n) = N$.
