33 Let $n$ be a positive integer, and let $d(n)$ denote the number of positive divisors of $n$, and $\varphi(n)$ denote the number of integers in $1,2, \cdots, n$ that are coprime to $n$. Find all $n$ such that $d(n)+\varphi(n)=n$.
33. Let $n$ be a number that satisfies the condition, then $n>1$, and at this time, only one number (i.e., 1) is both a divisor of $n$ and coprime with $n$. Therefore, among the numbers $1,2, \cdots, n$, there is exactly one number that is neither a divisor of $n$ nor coprime with $n$.

Since $\varphi(n)$ is even when $n>1$, if $n$ is even, then for $n \geqslant 10$, the numbers $n-2$ and $n-4$ are neither coprime with $n$ nor divisors of $n$, so $n \geqslant 10$ and $n$ being even does not satisfy the condition; if $n$ is odd, then $d(n)$ is odd, in which case $n$ is a perfect square. Let $n=(2 m+1)^{2}$, if $m>1$, then the numbers $n-(2 m+1)$ and $n-2(2 m+1)$ are neither divisors of $n$ nor coprime with $n$, at this time, $n$ does not satisfy the condition.

In summary, we only need to verify for $n=2,4,6,8,9$, and it can be known that the numbers satisfying the condition are $n=6,8,9$.