Let the total number of sheep be 10x+y, where y<10. The total money raised is (10x+y)^{2} = 100x^{2} + 20xy + y^{2}. Regardless of the values of x and y 100x^{2} + 20xy will be divisible by 20. Because the number of $10 bills is odd y^{2} mod 20 must be greater than 10 and less than 20. The only values of y where this is true is 4 and 6, where y^{2} is either 16 or 36. Either way there will be 6 $1 bills left over. Before the check the first brother will have $4 more than the second brother. A $2 check will balance things out.

Thanks to Bill Feldman for suggesting this problem.

Michael Shackleford, A.S.A.