As
$$\begin{align*} \min\{f(x_i), f(x_j)\} &= \min\{(x_i + a)(x_i + b), (x_j + a)(x_j + b)\} \\ &\le \sqrt{(x_i + a)(x_i + b)(x_j + a)(x_j + b)} \\ &\le \frac{1}{2}((x_i + a)(x_j + b) + (x_i + b)(x_j + a)) \\ &= x_i x_j + \frac{1}{2}(x_i + x_j)(a + b) + ab, \end{align*}$$
so
$$\begin{align*} F &\le \sum_{1 \le i < j \le n} x_i x_j + \frac{a+b}{2} \sum_{1 \le i < j \le n} (x_i + x_j) + C_n^2 \cdot ab \\ &= \frac{1}{2} \left[ \left( \sum_{i=1}^n x_i \right)^2 - \sum_{i=1}^n x_i^2 \right] + \frac{a+b}{2} (n-1) \sum_{i=1}^n x_i + C_n^2 \cdot ab \\ &= \frac{1}{2} \left( 1 - \sum_{i=1}^n x_i^2 \right) + \frac{n-1}{2} (a+b) + C_n^2 \cdot ab \\ &\le \frac{1}{2} \left( 1 - \frac{1}{n} \left( \sum_{i=1}^n x_i \right)^2 \right) + \frac{n-1}{2} (a+b) + C_n^2 \cdot ab \\ &= \frac{1}{2} \left( 1 - \frac{1}{n} \right) + \frac{n-1}{2} (a+b) + \frac{n(n-1)}{2} ab \\ &= \frac{n-1}{2} \left( \frac{1}{n} + a + b + nab \right). \end{align*}$$
The equality holds when $x_1 = x_2 = \cdots = x_n = \frac{1}{n}$. So the maximum of $F$ is $\frac{n-1}{2} \left( \frac{1}{n} + a + b + nab \right)$.

We show that the maximum value is attained when $x_1 = x_2 = \cdots = x_n = \frac{1}{n}$, and $F_{\max} = \frac{n-1}{2} \left(\frac{1}{n} + a + b + nab\right)$.
We induct on $n$ to show a more general statement: for non-negative real numbers $x_1, x_2, \dots, x_n$ satisfying $x_1 + x_2 + \dots + x_n = s$ (where $s$ is a fixed non-negative real number), the maximum value of $F = \sum_{1 \le i < j \le n} \min\{f(x_i), f(x_j)\}$ is attained when $x_1 = x_2 = \cdots = x_n = \frac{s}{n}$.
Since $F$ is symmetric, we may assume $x_1 \le x_2 \le \cdots \le x_n$. Note that $f(x)$ is strictly increasing on non-negative real numbers, we have
$$F = (n-1)f(x_1) + (n-2)f(x_2) + \cdots + f(x_{n-1}).$$
When $n=2$, $F = f(x_1) \le f(\frac{s}{2})$, equality holds when $x_1 = x_2$. Assume that the statement holds for $n$, consider the case of $n+1$. Applying inductive hypothesis on $x_2 + x_3 + \cdots + x_{n+1} = s - x_1$, we have
$$F \le nf(x_1) + \frac{1}{2}n(n-1)f(\frac{s-x_1}{n}) = g(x_1),$$
where $g(x)$ is a quadratic function of $x$, the leading coefficient is $1 + \frac{n-1}{2n^2}$, and the coefficient of $x$ is $a + b - \frac{n-1}{2n}(a + b + \frac{s}{2n})$, therefore, the axis of symmetry is
$$\frac{\frac{n-1}{2n}\left(a+b+\frac{s}{2n}\right)-a-b}{2+\frac{n-1}{n^2}} \le \frac{s}{2(n+1)}.$$
(The above inequality is equivalent to $[(n-1)s - 2n(n+1)(a+b)](n+1) \le 2s(2n^2 + n - 1)$; obviously, left-hand side $< (n^2-1)s <$ right-hand side.) Therefore, $g(\frac{s}{n+1})$ is the maximum of $g(x)$ on $[0, \frac{s}{n+1}]$. Thus, $F$ attains its maximum when $x_2 = x_3 = \cdots = x_{n+1} = \frac{s-x_1}{n} = \frac{s}{n+1} = x_1$, completing the solution. $\square$