The general two-user memoryless multiple-access channel, with common channel state information known to the encoders, has no single-letter solution which explicitly characterizes its capacity region. In this paper a binary "dirty" multiple-access channel (MAC) with "common interference", when the interference sequence is known to both encoders, is considered. We determine its sum-capacity, which equals to the capacity when full-cooperation between transmitters is allowed, contrary to the Gaussian case. We further derive an achievable rate region for this channel, by adopting the "onion-peeling" strategies which achieve the capacity region of the "clean" binary MAC. We show that the gap between the capacity region of the clean MAC and the achievable rate region of dirty MAC stems from the loss of the point-to-point binary dirty channel relative to the corresponding clean channel.