10. Transporting utility poles from a construction site by the roadside along a straight road in the same direction to plant them 500 m away on the roadside, plant one at the 500 m mark, and then plant one every 50 m along the roadside. Knowing that the transport vehicle can carry a maximum of 3 poles at a time, to complete the task of transporting and planting 20 poles, and returning to the construction site, the minimum total distance the transport vehicle must travel is $\qquad$ m.
10. 14000 .

Assuming the completion of transporting and planting 21 utility poles, 3 poles each time, let the round trip distance for the $k(k=1,2, \cdots, 7)$-th time be $a_{k}$. Then
$$
a_{1}=2 \times 600=1200,
$$

and $a_{k+1}=a_{k}+2 \times 150(k=1,2, \cdots, 6)$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 1200 and a common difference of 300.
Thus, $S_{7}=7 \times 1200+\frac{7 \times 6}{2} \times 300=14700(\mathrm{~m})$.
However, in reality, only 20 poles are transported, so there must be one time when 2 poles are transported, and the other 6 times, 3 poles each.

If the transportation of 2 poles is arranged on the $k(k=1,2, \cdots, 7)$-th time, then $a_{1}, a_{2}, \cdots, a_{k-1}$ remain unchanged, and $a_{k}, a_{k+1}, \cdots, a_{7}$ each decrease by $100 \mathrm{~m}$.
$$
\begin{array}{l}
\text { Hence } S_{7}(k)=S_{7}-100(8-k) \\
=13900+100 k .
\end{array}
$$

Obviously, when $k=1$, that is, the first time 2 poles are transported, and the rest each time 3 poles, the total distance is minimized at $14000 \mathrm{~m}$.