Let the set of all positive divisors of $n$ be $A$, and let the set of numbers in $1,2, \cdots, n$ that are coprime with $n$ be $B$. Since there is exactly one number $1 \in A \cap B$ in $1,2, \cdots, n$, we have
$d(n)+\varphi(n) \leqslant n+1$.
Thus, $c=0$ or 1.
(1) When $c=0$, from $d(n)+\varphi(n)=n$, we know that there is exactly one number in $1,2, \cdots, n$ that does not belong to $A \cup B$. If $n$ is even and $n > 8$, then $n-2, n-4$ do not belong to $A \cup B$, so $n$ does not satisfy the equation. If $n$ is odd, when $n$ is a prime or 1, $d(n)+\varphi(n)=n+1$ (which falls under case (2)); when $n$ is a composite number, let $n=pq, 1<p \leqslant q, p, q$ are both odd numbers. If $q \geqslant 5$, then $2p, 4p$ do not belong to $A \cup B$, so $n$ does not satisfy the equation.

In summary, only when $n \leqslant 8, n$ is even, or $n \leqslant 9, n$ is an odd composite number, does
$$d(n)+\varphi(n)=n.$$

Direct verification shows that $n$ can only be $6, 8, 9$.
(2) When $c=1$, from $d(n)+\varphi(n)=n+1$, we know that every number in $1,2, \cdots, n$ belongs to $A \cup B$. It is easy to see that, in this case, $n=1$ or a prime number meets the requirement. For the case when $n$ is even, as discussed above, if $n$ is even, then $n \leqslant 4$ (consider the number $n-2$); if $n$ is an odd composite number, let $n=pq, 3 \leqslant p \leqslant q, p, q$ are both odd numbers, in this case, $2p$ does not belong to $A \cup B$, which is a contradiction.
Direct verification shows that $n=4$ meets the requirement.
Therefore, the $n$ that satisfies $d(n)+\varphi(n)=n+1$ is 1, 4, or a prime number.