Question 657407
In order to maximize the sum of the remainders, we need a number, x, that gives:
(1) remainder 1 when divided by 2, or x = 2m+1
(2) remainder 2 when divided by 3, or x = 3n+2
(3) remainder 3 when divided by 4, or x = 4o+3
(4) remainder 4 when divided by 5, or x = 5p+4
where m, n, o, and p are integers >= 0

From (1) we know the number must be odd, so both n and p must be odd.  This gives us 

(2a) x = 6n + 5 and
(4a) x = 10p + 9

We can start listing possible values of x for each equation (2a), (3), and (4a):
(2a) 5, 11, 17, 23, 29, 35, 41, 47, 53, 59
(3)  3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59
(4a) 9, 19, 29, 39, 49, 59

59 is the smallest number that meets all the criteria, and the sum of the digits is 14.