Answer: 89

Small Hint:

With $c$ children the grape counts are an arithmetic sequence, and the total is $c(n + 1 - c) = 2009 = 7^2 \cdot 41$

Big Hint:

The last-place child ate $n + 2 - 2c \ge 0$ grapes; check which divisors $c$ of $2009$ survive that condition

Solution:

Let $c$ be the number of children. The grape counts $n,$ $n - 2,$ $\ldots,$ $n + 2 - 2c$ form an arithmetic sequence, so the total is $c$ times the average of the first and last terms: $$\begin{aligned} &c \cdot \frac{n + (n + 2 - 2c)}{2} \\ &= c(n + 1 - c) \\ &= 2009 = 7^2 \cdot 41. \end{aligned}$$ Thus $2009$ is divisible by $c$ and $n = \frac{2009}{c} + c - 1.$

The last-place child ate $n + 2 - 2c = \frac{2009}{c} + 1 - c \ge 0$ grapes, which forces $c(c - 1) \le 2009,$ ruling out $c = 49,$ $287,$ and $2009.$ The remaining divisors give $n = 2009$ for $c = 1,$ $n = 287 + 6 = 293$ for $c = 7,$ and $n = 49 + 40 = 89$ for $c = 41.$

The smallest possible value is $n = 89.$