$1$ and all odd prime numbers.

Clearly, $n = 1$ satisfies the condition. All odd primes also satisfy it: if $n = p$, then its divisors, increased by $1$, are $2$ and $p+1$; both divide $p+1$.

On the other hand, any number $n$ satisfying the condition has $1$ as a divisor; thus, $n+1$ is divisible by $1+1$, i.e., $n$ is odd.

Suppose now that some composite $n$ satisfies the condition. We have $n = ab$, where $a \geq b \geq 2$. Then $n+1$ is divisible by $a+1$; moreover, $n+b = (a+1)b$ is also divisible by $a+1$. Therefore, $b-1 = (n+b)-(n+1)$ is also divisible by $a+1$. Since $b-1 > 0$, we get $b-1 \geq a+1$. But this contradicts the inequality $b \leq a$.