You had the right idea, but this seems to be a fairly difficult problem. My solution depends upon a non-trivial result about the median of the gamma distribution. My answer is probably not that satisfying, since it seems unlikely that participants in a math contest would know this stuff off hand; It took me a fair amount of research to come up with this solution. Hopefully someone else will come up with a more elementary argument.
Edit: I've decided to offer a bounty on this question since I really want to see a simpler solution. I offer 100 points to anyone who is able to prove that
$$\int_{0}^{n}e^{-t}\left(1 + t + {t^2\over 2!} + \ldots + {t^{n} \over n!}\right)dt > {n\over 2}$$
for all positive integers $n$, in a simpler manner more consistent with the spirit of a math competition. I know this is a somewhat subjective requirement, what I'm looking for is an argument a bright first year calculus student could come up with without access to the internet or other research materials.
My Solution:
It was a good idea to define $G(x)=F(x)-{n \over 2}$. We can use the Fundamental Theorem of Calculus to compute the derivative of $G(x)$. This yields
$$G^{'}(x)=e^{-x}(1+x+{x^2 \over 2!}+{x^3 \over 3!}+\ldots + {x^n \over n!})$$
Claim: $G^{'}(x) > 0$ $\forall x \in [{n \over 2}, n]$
Proof
Since $e^{-x}$ is positive for all $x$, this depends on whether or not
$$p_n(x) = 1 + x +\ldots + {x^n \over n!} > 0$$. Since we are dealing with the interval $[{n\over 2}, n]$ for positive $n$, all terms of this sum are positive
for $x$ in our interval. Thus $p_n(x)>0$ for $x \in [{n \over 2}, n]$.
Thus we have shown that $G^{'}(x) > 0$ for $x \in [{n \over 2}, n]$.
$\square$
This shows that $G(x)$ is monotonically increasing. Now we look at
$G({n \over 2})$.
Claim: $G({n \over 2}) < 0$.
Proof.
Since $1 + x + \ldots + {x^n \over n!} < e^x$ for all positive $x \in \mathbb{R}$.
$$G({n \over 2}) = \int_{0}^{{n \over 2}}e^{-t}(1 + t +\ldots + {t^n \over n!})dt - {n \over 2}$$
$$< \int_{0}^{{n \over 2}}e^{-t}e^{t}dt - {n \over 2} < {n \over 2} - {n \over 2} = 0$$
$\square$
Next we consider $G(n)$.
Claim: $G(n)>0$.
Proof.
This is the hard part
$$G(n) = \int_{0}^{n}e^{-t}p_n(t)dt - {n \over 2}$$
$$p_n(x) = e^{x} - R_n(x) = e^{x} - \int_{0}^{x}{e^{t} \over n!}(x-t)^ndt$$
where we have applied Taylor's formula with the Cauchy form for the remainder.
Hence
$$G(n) = \int_{0}^n\left(1 - e^{-t}\int_{0}^{t}{e^{u}\over n!}(t - u)^ndu\right) dt - {n\over 2}$$
$$ = \int_{0}^n\left(1 - \int_{0}^{t}{e^{u-t}\over n!}(t - u)^ndu\right) dt - {n\over 2}$$
Making the change of variables $v = t - u$ in the inner integral we get
$$ G(n) = \int_{0}^n\left(1 - {1\over n!}\int_{0}^{t}e^{-v}v^ndv\right) dt - {n\over 2} = \int_{0}^{n}1 - {\gamma(n+1,t) \over n!}dt - {n \over 2}$$
Where $\gamma(s,x)$ is the lower incomplete gamma function. $\gamma(s,x)$ is complementary with $\Gamma(s,x)$, the upper incomplete gamma function.
$\gamma(s,x) + \Gamma(s,x) = \Gamma(s)$, the standard gamma function. Applying this fact yields
$$G(n) = {1\over n!}\int_{0}^{n}\Gamma(n+1,t)dt - {n \over 2}$$
Since $\Gamma(n+1,t)$ is a decreasing function in $t$ for fixed $n$ we have
$$G(n) > n{\Gamma(n+1,n)\over n!} - {n \over 2}$$
The integral is greater than the minimum value of the integrand in the interval in question times the length of the interval.
Now let's take a look at the Gamma Distribution. In particular we want to consider the distribution $Gamma(n+1,1)$ which has cumulative distribution function
$$f(x) = {\gamma(n+1,x) \over n!}.$$
Where $x \in (0,\infty)$.
Thus if $X$ is a random variable drawn from this distribution then
$$P(X < x) = {\gamma(n+1,x) \over n!}$$
We also have
$$P(X > x) = {\Gamma(n+1,x) \over n!}$$
Which follows from the complementary nature of the incomplete gamma functions.
According to this paper, the median $\nu$ of $Gamma(n+1,1)$ satisfies
$$n + {2 \over 3} < \nu < n + \ln (2)$$.
In particular this tells us that $n$ is less than the median of the distribution. Thus
$${\Gamma(n+1,n) \over n!} = P(X > n) > {1 \over 2}$$
which tells us that
$$G(n) > n{\Gamma(n+1,n) \over n!} - {n \over 2} > {n\over 2} - {n\over 2} = 0$$
$\square$
We have proven that $G(x)$ increases strictly monotonically on $[{n\over 2},n]$, that $G({n\over 2})$ < 0, and that $G(n) > 0$. Hence it follows that $G(x)$ has one and only root in this interval. Thus the equation
$$F(x) = {n\over 2}$$
has one and only one root in $[{n\over 2}, n]$.
