\(n;\ n+1;\ n+2;\ n+3-four\ consecutive\ numbers\\\
+(n+1)+(n+2)+(n+3)=410\
+n+1+n+2+n+3=410\\4n+6=410\ \ \ \ |subtract\ 6\ from\ both\ sides\\4n=404\ \ \ \ |divide\ both\ sides\ by\ 4\\\boxed{n=101}\\\boxed{n+1=102}\\\boxed{n+2=103}\\\boxed{n+3=104}\\\\Answer:\boxed{\boxed{101;\ 102;\ 103;\ 104}}\)