Go to the floor

$\left \lfloor{\sqrt{1}}\right \rfloor=1$

$\left \lfloor{\sqrt{4}}\right \rfloor=2$

$\left \lfloor{\sqrt{9}}\right \rfloor=3$

$\left \lfloor{\sqrt{16}}\right \rfloor=4$

$\left \lfloor{\sqrt{25}}\right \rfloor=5$

$\left \lfloor{\sqrt{36}}\right \rfloor=6$

$\left \lfloor{\sqrt{49}}\right \rfloor=7$

$\left \lfloor{\sqrt{64}}\right \rfloor=8$

$\left \lfloor{\sqrt{81}}\right \rfloor=9$

$\left \lfloor{\sqrt{100}}\right \rfloor=10$

Note that our floor function will change its value each square number so we have to sustract consecutive squares to find the intervals that contain each value.

$4-1=3$ numbers whose value is $1$

$9-4=5$ numbers whose value is $2$

$16-9=7$ numbers whose value is $3$

$25-16=9$ numbers whose value is $4$

$36-25=11$ numbers whose value is $5$

$49-36=13$ numbers whose value is $6$

$64-49=15$ numbers whose value is $7$

$81-64=17$ numbers whose value is $8$

$100-81=19$ numbers whose value is $9$

and $\left \lfloor{\sqrt{10}}\right \rfloor=10$

$\therefore$ the value is $1(3)+2(5)+3(7)+4(9)+...+8(17)+9(19)+10=\boxed{625}$

Java solution -

double n = 0;
for(int i = 1; i <= 100; i++)
    n += Math.floor(Math.sqrt(i));
System.out.println(n);

Python:

1
2

from math import floor, sqrt
print sum([floor(sqrt(i)) for i in xrange(1,101)])