I've run out of problem titles!

Obviously, since $120$ and $n$ are both integers, they can be factored in the form $a \cdot b \cdot c \ldots$ for factors $a, b, c, \ldots$ .

Now, since $\phi(p^k) = p^{k-1} \cdot (p-1)$ for primes $p$ , we should factor 120 into the form $(p_a^{k_a-1} \cdot (p_a-1)) \cdot (p_b^{k_b-1} \cdot (p_b-1) \ldots$ for some primes $p_a, p_b \ldots$ . Then, we get $n = p_a^{k_a} \cdot p_b^{k_b} \ldots$ .

Now, since $n$ is to be minimized, we can get the minimum value of $n$ if there is only one prime factor of $n$ , and $k=1$ , since this gives us $n = p$ for some prime p, and $p-1 = 120$ . But, this implies that $p=121,$ , and since $121$ is not prime, we reject this solution.

Our next smallest value of $n$ can be acquired f there is only one prime factor of $n$ , and $k=2$ , since this implies that $n = p^2$ for some prime p, and $(p)(p-1) = 120$ . But, since 120 cannot be factored into two consecutive integers, this idea is to be tossed into the trash bin.

Our next smallest value of $n$ can be acquired when there are two prime factors of $n$ , namely $p_a, p_b$ and $k_a = k_b = 1$ , since this implies that $n = p_a \cdot p_b$ and $(p_a-1) \cdot (p_b-1) = 120$ . Also, $|p_a-p_b|$ must be minimized in order to minimize $n$ . By factoring out $120$ , we get that $120 = 12 \cdot 10 = (13-1)(11-1)$ . This implies that $n = 13 \cdot 11 = \boxed {143}$ , which is our final answer.