, 2012

The answer is $n = 1$, $n = 4$, or $n$ is any prime number. Let $S$ be the set of integers less than $n$ and relatively prime to $n$, and let $T$ be the set of integers that are positive divisors of $n$. Then
$$(i) \quad |S| = \phi(n) \text{ and } |T| = \tau(n),$$
$$(ii) \quad S, T \subseteq \{1, 2, \dots, n\}, \text{ and}$$
$$(iii) \quad S \cap T = \{1\}.$$
We require $|S| + |T| > n$. By the principle of inclusion-exclusion, $|S| + |T| - |S \cap T| = |S \cup T|$. We know $|S \cap T| = 1$, and by $S, T \subseteq \{1, 2, \dots, n\}$, we have $|S \cup T| \le n$. Therefore $|S| + |T| > n$ if and only if $|S \cup T| = n$, meaning that every positive integer in $\{1, 2, \dots, n\}$ is either relatively prime to $n$ or a divisor of $n$. It is easy to check that this is true for $n = 1$, $n$ prime, and $n = 4$. Now suppose $n$ is a composite number.

Let $d$ be the smallest divisor of $n$ greater than $1$. $d < n$ since $n$ is not prime. Consider $n-d$. We have $\gcd(n-d, n) = d > 1$, so $n-d$ must be a divisor of $n$. But $n = k d$ for some $k$, so if $n-d$ divides $n$ then we have an $m$ such that $m(kd - d) = k d$. This gives $m(k-1) = k$. This is only possible for $k=2$, so $n = 2d$.

Then since $n$ is even, $2$ is the smallest divisor of $n$ greater than $1$ and $d = 2$. Therefore $n = 4$, and this is the only composite number that works.