
simulate this circuit – Schematic created using CircuitLab
If I read correctly, the poster is doing this, and is experiencing voltage drop due to the length and resistance of the wire.
The reason for the voltage drop is because the wire has resistance, and you are drawing high current (350mA) for each series string.
The \$1\space\text{mm}^2\$ wire has a resistance of approx. 20 ohms/kilometer, so it is 0.020 ohms / meter. Assuming that you have the LED strings equally spaced across 200 meters, then your circuit actually looks like this, electrically:

simulate this circuit
At each point along the wire, you are increasing the wire resistance, and requiring that wire to take more and more current. As an example, take the LED string in the middle, represented by I3.
At that node, I3, in my picture above, there is 6 x 0.666 = 4 ohms back to the source voltage, and that 4 ohms of resistance is carrying the current of I3, I4, I5 and I6 = 350mA x 4 = 1.4 Amps ( and that's a simplification, as I neglect the current and voltage drops contributed by the LEDs before it, at I1 and I2 -- forget about them for now) .. so that 1.4A across 4 Ohms causes a voltage drop of 1.4 x 4 = 5.6 Volts, according to ohms law.
So you see, there is a voltage drop at that mid point of 5.6V or more, and this is too low to light the remaining LEDs (I4, I5 and I6).
The simple solution is to raise the input voltage, if your LED can handle it, or increase the thickness of the wire to maybe \$4\space\text{mm}^2\$ wire. This will have much less resistance, dropping only a few volts along the 200m length.
The proper and correct solution is to raise the voltage to about 18-20V, use thicker wire, maybe \$2.5 - 4\space\text{mm}^2\$, and use a constant current regulation for 350mA at each string. This is the only way to get each string to be equal brightness.