Example 13 (2004 Western China Mathematical Olympiad) Let $n \in \mathbf{N}_{+}$, and let $d(n)$ denote the number of all positive divisors of $n$, and $\varphi(n)$ denote the number of integers in $1,2, \cdots, n$ that are coprime to $n$. Find all non-negative integers $c$ such that there exists a positive integer $n$ satisfying $d(n)+\varphi(n)=n+c$, and for each such $c$, find all positive integers $n$ that satisfy the equation.
Let the set of all positive divisors of $n$ be $A$, and 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 $d(n)+\varphi(n) \leqslant n+1$, hence $c=0$ 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 $A \cup B$, at this time $n$ does not satisfy the equation; If $n$ is odd, then when $n$ is a prime or 1, $d(n)+\varphi(n)=n+1$ [which belongs to case (2)], when $n$ is a composite number, let $n=p q, 1<p \leqslant q, p, q$ are both odd numbers, if $q \geqslant 5$, then $2 p, 4 p$ do not belong to $A \cup B$, at this time $n$ 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 $d(n)+\varphi(n)=n$ hold. Direct verification shows that: $n$ can only be 6, 8, and 9.
(2) When $c=1$, from $d(n)+\varphi(n)=n+1$ we know that each 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 where $n$ is even (not a prime), as discussed above, $n \leqslant 4$ (consider the number $n-2$); if $n$ is an odd composite number, let $n=p q, 3 \leqslant p \leqslant q, p, q$ are both odd numbers, at this time $2 p \notin A \cup B$, 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.
Note: $A \cap B=\{1\}$ is obvious, hence $c \leqslant 1$. When $n$ is a prime, it is easy to see that $d(n)+\varphi(n)=n+1$; when $n$ is a composite number, for sufficiently large composite numbers $n$, it is easy to find a sufficient number of composite numbers less than $n$ that are neither divisors of $n$ nor coprime with $n$, at this time $d(n)+\varphi(n) \neq n+c (c=0$ or 1$)$. The general case of this problem is to remove the restriction that the integer $c$ is "non-negative".