4. B. Given positive integers $a_{1}, a_{2}, \cdots, a_{10}$ satisfying
$$
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{10} \text {, }
$$

and no three of these numbers can be the lengths of the sides of a triangle. Then the minimum value of $\frac{a_{10}}{a_{1}}$ is ( ).
(A) 34
(B) 55
(C) 89
(D) 144
4. B. B.

Notice
$$
\begin{array}{l}
a_{10} \geqslant a_{9}+a_{8} \geqslant 2 a_{8}+a_{7} \geqslant 3 a_{7}+2 a_{6} \\
\geqslant 5 a_{6}+3 a_{5} \geqslant 8 a_{5}+5 a_{4} \geqslant \cdots \\
\geqslant 34 a_{2}+21 a_{1} \geqslant 55 a_{1} . \\
\text { Therefore, } \frac{a_{10}}{a_{1}} \geqslant 55 .
\end{array}
$$

When $a_{1}=a_{2}, a_{n+2}=a_{n+1}+a_{n}(1 \leqslant n \leqslant 8)$, the equality holds.
Thus, the minimum value of $\frac{a_{10}}{a_{1}}$ is 55.