Let $a_1, a_2, \ldots, a_n$ be a permutation of $1, 2, \ldots, n$. We aim to find the minimum of
$$\sum_{i=1}^n \min \{ a_i, 2i-1 \}.$$

We claim that the minimum is achieved when $a_i = n + 1 - i$ for all $i$. In this configuration, the terms $b_i = \min(a_i, 2i-1)$ will be structured as follows:
- For $i$ from $1$ to $\left\lfloor \frac{n+2}{3} \right\rfloor$, $b_i = 2i-1$.
- For $i \geq \left\lceil \frac{n+2}{3} \right\rceil$, $b_i = n + 1 - i$.

In the sequence $b_i$, which ranges from $1$ to $n + 1 - \left\lceil \frac{n+2}{3} \right\rceil$, each odd number up to the upper bound appears twice, and each even number up to the upper bound appears once.

To show that this is indeed the minimum, note that each odd number can appear at most twice (once as an $a_i$ and once as $2i-1$), and each even number can appear only once (as an $a_i$). Therefore, the minimum is achieved by greedily taking all the smaller numbers, i.e., two 1s, one 2, two 3s, and so on, which aligns with the described configuration.

Thus, the minimum value of the sum is:
$$\sum_{i=1}^n \min \{ a_i, 2i-1 \}.$$
The answer is: \boxed{\sum_{i=1}^n \min \{ n + 1 - i, 2i-1 \}}.