1000 roses

Keep in mind that Agnis has to give out an integral number of roses to an integral number of girls, and he must give strictly more roses at each subsequent girl. To give roses to the most girls, Agnis would have to distribute the roses in fashion of $1, 2, 3 ,4... n$ . So, Agnis is giving out $1+2+3+4+...+n$ roses. So essentially, we're looking for the largest triangular number $T_n$ such that $T_n = \frac{n(n+1)}{2} < 1000$ , where $n$ is our answer, which happens to be $\boxed{44}$ .

Using the formula for the sum of the Arithmetic Progression to get the maximum value of n:

$S_n= \frac{n}2(2a_1+(n-1)d)$

$Where: S_n=1000$ $a_1=1$ $d=1$

We get:

$2000=n(2(1)+n-1) \rightarrow n^2+n-2000=0$

Using the quadratic formula to solve for the positive value of n:

$n=\frac{-1+\sqrt{1-(4)(-2000)(1)}}{2(1)}$ $n=\frac{-1+\sqrt{8001}}{2}$ $n \approx 44.74 \ldots= \boxed{44}$

Since we are talking to a person (n should be an integer), n can't be 44.74, so the maximum number of girls is 44.

A little Python simulation will do just fine.

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# AGNIS'S LOVE MACHINE v1.0
roses = 1000
giving_away = 1
girls = 0
# while he still has more to give
while giving_away <= roses:
    # give away roses
    roses -= giving_away
    girls += 1
    # give one more rose next time
    giving_away += 1
print "Answer:", girls

Trial and Error n(n-1) all over 2. sqrt(1000)=10 sqrt(10)