3. Write 10 distinct positive integers $a_{1}, a_{2}, \cdots, a_{10}$ in the form of the following chart, where the number pointed to by two arrows equals the sum of the two numbers at the starting points of these arrows. For example $\stackrel{a_{1} \rightarrow a_{2}}{a_{5}}$ indicates $a_{2}=a_{1}+a_{5}$. Then, the smallest possible value of $a_{4}$ that satisfies the chart on the right is
A. 10
B. 20
C. 30
D. 40
3. B $a_{4}=a_{3}+a_{7}=\left(a_{2}+a_{6}\right)+\left(a_{6}+a_{9}\right)=a_{2}+2 a_{6}+a_{9}=\left(a_{1}+a_{5}\right)+2\left(a_{5}+a_{8}\right)+\left(a_{8}+\right.$ $\left.a_{10}\right)=a_{1}+3 a_{5}+3 a_{8}+a_{10}$. To make $a_{4}$ the smallest, we can first consider the values of $a_{5}$ and $a_{8}$, then consider the values of $a_{1}$ and $a_{10}$. (1) First, the values of $a_{5}$ and $a_{8}$ should be as small as possible. By symmetry, let's take $a_{5}=1, a_{8}=2$, then $a_{6}=3$. (2) Second, the values of $a_{1}$ and $a_{10}$ should be as small as possible. By symmetry, let's take $a_{1}=4$ (the case for $a_{10}=4$ can be similarly deduced), at this time $a_{2}=5, a_{3}=8$. When $a_{10}=6$, $a_{9}=8$, which contradicts $a_{3}=8$; when $a_{10}=7$, $a_{9}=9$, and $a_{7}=12$, thus $a_{4}=20$. Therefore, the minimum possible value of $a_{4}$ is 20.