Answer is $30$.Let's prove that it can't be done in less then $30$ moves. Assume that one did it in $x$ moves,$x<30$, after $x$-th move we have at least $100-x\geq 70$ children that must have over $100$ candies (because they weren't sharing candies),because all of them have different number of candies,that set of children got at least $0+1+2+3+...+70=\frac{70\cdot 71}{2}=2494$, but then the set of children that distributed the candies have less then $10000-100\cdot 71-2494=406$ but $1+2+3+...+29=435$,contradiction! Algorithm for $30$ is easy,Let first kid give $100-2$ candies to second,second give $198-3$ candies to third,... $29$th kid give $2900-(2+3+..+29+30)=2436$ candies so 29th kid have $2536$ candies,enough to left himself only one candie and give $1$ to the $32$th ,$2$ to $33th$..$68$ to the $99$th kid and the rest (greater then $68$) to the $100$th kid.