Decimal Roots

Note that

$({ n }^{ 2 }+n)<{ (n+0.5) }^{ 2 }=({ n }^{ 2 }+n+0.25)<({ n }^{ 2 }+n+1)$

Therefore all numbers k such that ${ n }^{ 2 }\le k\le ({ n }^{ 2 }+n)$ will have decimal parts that are less than 0.5; in other words, for up to 961, there will be

$(31+\frac { 930 }{ 2 } =496$ ) numbers of this form. From 961 to 1000, there are another 31 numbers of this form. Therefore the answer is

$100(496+31)=100(527)=\boxed { 52700 }$