2. For any $m \in \mathbf{N}_{+}$, try to find the sum $s=\sum_{d \mid m} \varphi(d)$.


2. Solution: Let $A=\{1,2, \cdots, m-1, m\}$, for $d \mid m, d \in \mathbf{N}_{+}$.

Let the positive integers not greater than $d$ and coprime with $d$ be $a_{1}, a_{2}, \cdots, a_{\varphi(d)}$, then it is easy to know that $\frac{m}{d} \cdot a_{1}, \frac{m}{d} a_{2}, \cdots$, $\frac{m}{d} a_{\varphi(d)}$, are all the positive integers in $s$ whose greatest common divisor with $m$ is $\frac{m}{d}$. When $d$ takes all divisors of $m$, $\frac{m}{d}$ also takes all divisors of $m$. Thus, each number in $s$ is counted exactly once in the sum of all $\varphi(d)$.
Therefore, $s=\sum_{d \mid m} \varphi(d)=m$.