As a mere arithmetical problem this question presents no difficulty. In order that the hands shall all point to twelve o'clock at the same time, it is necessary that B shall gain at least twelve hours and that C shall lose twelve hours. 

As B gains a minute in a day of twenty-four hours, and C loses a minute in precisely the same time, it is evident that one will have gained 720 minutes (just twelve hours) in 720 days, and the other will have lost 720 minutes in 720 days. 
Clock A keeping perfect time, all three clocks must indicate twelve o'clock simultaneously at noon on the 720th day from April 1, 1898. 

What day of the month will that be?

I published this little puzzle in 1898 to see how many people were aware of the fact that 1900 would not be a leap year. 
It was surprising how many were then ignorant on the point. 
Every year that can be divided by four without a remainder is bissextile or leap year, with the exception that one leap year is cut off in the century. 1800 was not a leap year, nor was 1900. 
On the other hand, however, to make the calendar more nearly agree with the sun's course, every fourth hundred year is still considered bissextile. 
Consequently, 2000, 2400, 2800, 3200, etc., will all be leap years. May my readers live to see them. 
We therefore find that 720 days from noon of April 1, 1898, brings us to noon of March 22, 1900.