Answer     13

Trying a few small numbers finds that it appears $\boxed{13}$ could be a candidate. Let's prove this.

Let's write $n$ in the form of $2^k\cdot m$ where $k$ is a non-negative integer and $m$ is an odd integer.

If $k \ge 4$, then $$\varphi(n) = \varphi(2^k)\varphi(m) \ge \varphi(2^k) = 2^{k-1} \ge 8$$

Therefore, it is sufficient to just further investigate those cases where $k=0, 1, 2, 3$.

When $k=0$, then $n$ is an odd.

• If $n$ is a prime, that $\varphi(n) = n-1 \ge 13-1 = 12$.
• If $n$ divides two distinct prime numbers $p > q$, then we must have $p\ge 5$ and $q\ge 3$. Hence, $$\varphi(n) \ge \varphi(pq) =\varphi(p)\varphi(q) = (p-1)(q-1) \ge 8$$
• If $n$ divides a square of prime number $p$, then $p \ge 3$. It follows that $$\varphi(n) \ge \varphi(p^2) = p(p-1) \ge 6$$

This case also means that for any odd number $n$, $\varphi(n) \ge \min(n-1, 6)$. We will reuse this conclusion.

When $k=1$, then $n=2 m$ where $m$ is an odd integer greater than $\lfloor{13/2}\rfloor = 6$. Then we have $$\varphi(n)=\varphi(2)\varphi(m)=\varphi(m) \ge \min(m-1, 6) \ge 6$$

When $k=2$, then $n=4m$ where $m$ is odd and $m > \lfloor{13/4}\rfloor = 3$. Therefore  $$\varphi(n)=\varphi(4)\varphi(m)=2\varphi(m)\ge 2\times \min(m-1, 6) \ge 2\times 3 = 6$$

When $k=3$, then $n=8m$ where $m > \lfloor{17/8}\rfloor = 2$. Because $m$ is odd, we find $m\ge 3$. Therefore $$\varphi(n) =\varphi(8)\varphi(m) = 4\varphi(m) \ge 4\times \min(m-1, 6)\ge 4\times 2=8$$

Hence, we conclude $13$ is the desired answer.