设n个连续的自然数为a+1,a+2,a+3,…,a+n,
则它们的和为:(a+1)+(a+2)+(a+3)+…+(a+n)=2006
即:na+
=2006
当n=4时,a=499,所以500+501+502+503=2006.
当n=17时,a=109,所以110+111+112+113+114+115+116+117+118+119+120+121+122+123+124+125+126=2006.
故答案为:(1)500+501+502+503,(2)110+111+112+113+114+115+116+117+118+119+120+121+122+123+124+125+126.