Example 6 Find the positive integer $n$ that satisfies the following conditions: $\varphi(n)>2$, and for any positive integers $a>b>c$ not exceeding $n$, if $(a, n)=(b, n)=(c, n)=1$, then $(a+b+c, n)=1$.
【Analysis】 $\varphi(n)>2$, so $n=5$ or $n \geqslant 7$.
When $n=5$, by $1+2+3+4=10$, we know that the sum after removing any number is coprime with 5, so $n=5$ satisfies the condition.

When $n=7$, by $(1+2+4,7) \neq 1$, we know the proposition does not hold.

When $n=8$, by $1+3+5+7=16$, we know that the sum after removing any number is coprime with 8, so $n=8$ satisfies the condition.

When $n=9$, by $(1+4+7,9) \neq 1$, we know the proposition does not hold.

When $n=10$, by $1+3+7+9=20$, we know that the sum after removing any number is coprime with 10, so $n=10$ satisfies the condition.

When $n=11$, by $2+3+6=11$, we know the proposition does not hold.

When $n=12$, by $1+5+7+11=24$, we know that the sum after removing any number is coprime with 12, so $n=12$ satisfies the condition.

When $n=13$, by $2+4+7=13$, we know the proposition does not hold.

When $n=14$, by $1+9+11=21$, we know the proposition does not hold.

When $n=15$, by $1+4+7=12$, we know the proposition does not hold.
When $n=16$, by
$(a, n)=(b, n)=(c, n)=1$,

we know these three numbers are all odd, and their sum is also odd, which is obviously coprime with 16, so $n=16$ satisfies the condition.

In summary, when $n \leqslant 16$, $n=5,8,10,12,16$ satisfy the condition.

From the proof that $n=8,16$ satisfy the condition, we know that $n=2^{k}$ $(k \geqslant 3)$ all satisfy the condition. It is easy to give the following proof.
When $n=2^{k}(k \geqslant 3)$, by
$(a, n)=(b, n)=(c, n)=1$,

we know $a, b, c$ are all odd, thus, $a+b+c$ is also odd, at this time, $(a+b+c, n)=1$, i.e., the proposition holds.
Below, we discuss the case when $n>16$.
Assume $n$ satisfies the condition.
Assume $n$ is odd, at this time,
$(2, n)=1,(4, n)=1,(8, n)=1,(16, n)=1$.
Thus, $(1+4+16, n)=1$, i.e.,
$(21, n)=1,(3, n)=1,(n-3, n)=1$.
At this time, $(1+2+(n-3), n)=n \neq 1$, contradiction.
Therefore, the assumption does not hold, i.e., when $n>16$, $n$ must be even.

Below, we classify and discuss $n$ based on modulo 3.
(1) $(3, n)=1$.
At this time, $(1, n)=1,(3, n)=1,(9, n)=1$
$\Rightarrow(13, n)=1$
$\Rightarrow(3+9+13, n)=1$
$\Rightarrow(5, n)=1 \Rightarrow(15, n)=1$.
At this time, $(1+5+15, n)=1$, then $(7, n)=1$;
$(1+3+7, n)=1$, then $(11, n)=1$;
$(1+3+9, n)=1$, then $(13, n)=1$.
Using mathematical induction, we can prove that: all odd numbers not exceeding $n$ are coprime with $n$, thus, we can conclude that $n$ can only be of the form $2^{k}$.
(2) $2|n, 3| n$, i.e., $n=6 k$.
At this time, it includes the following six cases.
(i) $k \equiv 0(\bmod 6)$, at this time, $1, k+1, 2 k+1$ are all coprime with $n$, and their sum $3 k+3$ is not coprime with $n$, which does not meet the condition.
(ii) $k \equiv 1(\bmod 6)$, at this time, $k-2, k+4, 3 k-2$ are all coprime with $n$, and their sum $5 k$ is not coprime with $n$, which does not meet the condition.
(iii) $k \equiv 2(\bmod 6)$, at this time, $k-1, 2 k+3, 3 k+1$ are all coprime with $n$, and their sum $6 k+3$ is not coprime with $n$, which does not meet the condition.
(iv) $k \equiv 3(\bmod 6)$, at this time, $k-2, k+4, 4 k+1$ are all coprime with $n$, and their sum $6 k+3$ is not coprime with $n$, which does not meet the condition.
(v) $k \equiv 4(\bmod 6)$, at this time, $1, k-3, 2 k-1$ are all coprime with $n$, and their sum $3 k-3$ is not coprime with $n$, which does not meet the condition.
(vi) $k \equiv 5(\bmod 6)$, at this time, $1, k-4, 2 k+3$ are all coprime with $n$, and their sum $3 k$ is not coprime with $n$, which does not meet the condition.

Combining (i) to (vi), we know that when $21 n, 31 n$, there are no $n$ that satisfy the condition.
Therefore, the $n$ that satisfy the condition can only be
$5,10,12,2^{k}(k \geqslant 3)$.