Negative Inputs For Diode Transistor Logic
That circuit looks like a NAND gate to me - Basically what happens is that if any of the inputs are driven low, it will pull down the voltage reverse biasing the pair of diodes meaning the transistor will be pulled down to zero by R2. If both of the inputs are high, the input diodes will not conduct and the transistor base will be pulled high through R1 causing it to switch on:

╔═══╦═══╦═════╦═══╗
║ A ║ B ║ Tr. ║ C ║
╠═══╬═══╬═════╬═══╣
║ 0 ║ 0 ║  0  ║ 1 ║
║ 0 ║ 1 ║  0  ║ 1 ║
║ 1 ║ 0 ║  0  ║ 1 ║
║ 1 ║ 1 ║  1  ║ 0 ║
╚═══╩═══╩═════╩═══╝


If the transistor is on, it will pull the output down to ground, hence the NAND function. Essentially the extra transistor acts as an inverter and hence you see the opposite polarity.

Interestingly if you invert both inputs what you will actually end up with is an OR gate, not an AND as you thought.



Edit 2

Based on the comments, it's now become clear why you were getting different results from what you expected. Basically the output is a current sink when low. As a result if you connect an LED from VCC to the transistor, the LED will be on when the output is low and off when the output is high. This is the case with any load wired in such a way, because only when the transistor is pulling down low is there any voltage across the load.

(Edit: I was right the first time, it's a NAND)