It is not possible to move all 16 bars.
Consider the total distance that each person moves, in each direction, carrying each possible number of bars. For example $R_2^+$ is the total distance that Rod moves forward while carrying 2 bars. We can set up a system of equations. Both Rod and Lia must move a net distance of $1$ mile forwards:
$$
R_0^+ - R_0^- + R_1^+ - R_1^- + R_2^+ - R_2^- = 1 \\
L_0^+ - L_0^- + L_1^+ - L_1^- = 1
$$
Both of them must spend less than $370$ minutes moving (note we have taken the reciprocals of their speeds in minutes per mile):
$$
20(R_0^+ + R_0^-) + 30(R_1^+ + R_1^-) + 60(R_2^+ + R_2^-) \le 370 \\
30(L_0^+ + L_0^-) + 40(L_1^+ + L_1^-) \le 370
$$
Finally, the total net distance moved by all the bars must be $16$ miles forwards:
$$
R_1^+ - R_1^- + 2(R_2^+ - R_2^-) + L_1^+ - L_1^- = 16
$$
And of course each distance must be nonnegative.
Next (thanks to RobPratt) we multiply these equations by $\frac{1}{2}$, $\frac{3}{7}$, $\frac{1}{40}$, $\frac{1}{70}$, and $-1$ respectively, and sum them, yielding:
$$
\phantom{+}\left(\dfrac{1}{2}+\dfrac{20}{40}\phantom{-0}\right)R_0^+
+ \left(-\dfrac{1}{2}+\dfrac{20}{40}\phantom{-(-0)}\right)R_0^- \\
+ \left(\dfrac{1}{2}+\dfrac{30}{40}-1\right)R_1^+
+ \left(-\dfrac{1}{2}+\dfrac{30}{40}-(-1)\right)R_1^- \\
+ \left(\dfrac{1}{2}+\dfrac{60}{40}-2\right)R_2^+
+ \left(-\dfrac{1}{2}+\dfrac{60}{40}-(-2)\right)R_2^- \\
+ \left(\dfrac{3}{7}+\dfrac{30}{70}\phantom{-0}\right)L_0^+
+ \left(-\dfrac{3}{7}+\dfrac{30}{70}\phantom{-(-0)}\right)L_0^- \\
+ \left(\dfrac{3}{7}+\dfrac{60}{70}-1\right)L_1^+
+ \left(-\dfrac{3}{7}+\dfrac{60}{70}-(-1)\right)L_1^- \\
\le \dfrac{1}{2} + \dfrac{3}{7} + \dfrac{370}{40} + \dfrac{370}{70} - 16
$$
Which simplifies to:
$$
R_0^+
+ \frac{1}{4}R_1^+
+ \frac{5}{4}R_1^-
+ 3R_2^-
+ \frac{6}{7}L_0^+
+ \frac{8}{7}L_1^-
\le -\frac{15}{28}
$$
Since the left hand is a sum of nonnegative terms, it must be nonnegative; but a nonnegative number cannot be less than or equal to a negative number. Therefore the system of equations has no solution, and the puzzle has no solution for 16 bars.
If the number of bars is reduced to 15, the right-hand side of the final equation increases by one and becomes positive, meaning that a solution for 15 bars is not ruled out. The maximum amount of extra time available is:
$$\frac{-\frac{15}{28}+1}{\frac{1}{40}+\frac{1}{70}}=\frac{\frac{13}{28}}{\frac{11}{280}}=\frac{130}{11}$$
Therefore the minimum time required is:
$$370-\frac{130}{11}=\frac{3940}{11}=5\text{h}58\tfrac{2}{11}$$
To achieve this minimum time all the terms on the left-hand side of the equation must be zero. Also, the inequalities will become equalities. This yields the following system of equations:
$$
- R_0^- + R_2^+ = 1 \\
- L_0^- + L_1^+ = 1 \\
20R_0^- + 60R_2^+ = \frac{3940}{11} \\
30L_0^- + 40L_1^+ = \frac{3940}{11} \\
2R_2^+ + L_1^+ = 15
$$
This system of five equations in four variables has a unique solution:
$$
R_0^- = \frac{41}{11},\
R_2^+ = \frac{52}{11},\
L_0^- = \frac{50}{11},\
L_1^+ = \frac{61}{11}
$$
Since the distances are not integers, the thieves must drop the bars off somewhere in the middle. Their movements will look like this:
Lia (gold line) carries 1 bar all the way to the station, then carries 12 bars from the midway point to the station in 12 trips. Rod (blue line) carries 12 bars to the midway point in 6 trips and then carries 2 bars all the way to the station. The total number of bars moved is 15: the one Lia carries plus the 2 Rod carries plus the 12 they both carry.
Rod makes 6 return trips with a total distance of $R_0^-=\frac{41}{11}$, so the midway point is $\frac{41}{66}$ miles from the start. Lia makes 12 return trips with a total distance of $L_0^-=\frac{50}{11}$, so the midway point is $\frac{25}{66}$ miles from the end. $\frac{41}{66}+\frac{25}{66}=1$ so this checks out. The total time taken by Rod is $6\left(60+20\right)\frac{41}{66}+60=\frac{3940}{11}$ and the total time taken by Lia is $40+12\left(30+40\right)\frac{25}{66}=\frac{3940}{11}$, which also checks out. Finally, looking at the plot, each of the times that Rod drops off a pair of bars comes before the two corresponding times when Lia picks up the bars.
