• $I_n=\{ m \in \mathbb{Z}: 1 \le m \le n \}$
• $D_n=\{ d \in I_n: d>1$ and $d|n \}$
• $T_n=\{ t \in I_n: t$ is a totative of $n \}$

Then $D_n \cup T_n=I_n$ if and only if $n=1$ $n=4$ or $n$ is prime.

Necessity:

If $n=1$ then $D_n=\emptyset$ and $T_n=\{1\}$ Thus, $D_n \cup T_n=I_n$

If $n=4$ then $D_n=\{2,4\}$ and $T_n=\{1,3\}$ Thus, $D_n \cup T_n=I_n$

If $n$ is prime, then $D_n=\{ n \}$ and $T_n=I_n \setminus \{ n \}$ Thus, $D_n \cup T_n=I_n$

Sufficiency:

This will be proven by considering its contrapositive.

Suppose first that $n$ is a power of $2$ Then $n \ge 8$ Thus, $6 \in I_n$ On the other hand, $6$ is neither a totative of $n$ (since $\gcd(6,n)=2$ nor a divisor of $n$ (since $n$ is a power of $2$ . Hence, $D_n \cup T_n \neq I_n$

Now suppose that $n$ is even and is not a power of $2$ Let $k$ be a positive integer such that $2^k$ exactly divides $n$ Since $n$ is not a power of $2$ it must be the case that $n=2^kr$ for some odd integer $r \ge 3$ Thus, $n=2^kr>2^{k+1}$ Therefore, $2^{k+1} \in I_n$ On the other hand, $2^{k+1}$ is neither a totative of $n$ (since $n$ is even) nor a divisor of $n$ (since $2^k$ exactly divides $n$ . Hence, $D_n \cup T_n \neq I_n$

Finally, suppose that $n$ is odd. Let $p$ be the smallest prime divisor of $n$ Since $n$ is not prime, it must be the case that $n=ps$ for some odd integer $s \ge 3$ Thus, $n=ps>2p$ Therefore, $2p \in I_n$ On the other hand, $2p$ is neither a totative of $n$ (since $\gcd(2p,n)=p$ nor a divisor of $n$ (since $n$ is odd). Hence, $D_n \cup T_n \neq I_n$