Answer: The minimum value is $\frac{k(4k^2 + k - 1)}{2}$.

Let us number each day of the tournament in order as $1, 2, \dots, C_2^{2k}$. Let us sort the arrival dates of each player as $b_1 \le b_2 \le b_{2k}$, and the departure dates as $e_1 \ge e_2 \ge \dots \ge e_{2k}$ (note that here this is not indexed by competitor number; a competitor may check in on day $b_i$ but check out on day $e_j$.) The total hotel cost is
$$\sum_{i=1}^{2k} (e_i - b_i + 1).$$

(1) Estimation: For this, we need to estimate $e_i - b_i + 1$. Note that before day $b_{i+1}$ only $i$ competitors have checked in, so there can be at most $C_2^i$ matches, thus $b_{i+1} \le C_2^i + 1$. Similarly, on day $e_{i+1}$ only $i$ competitors remain, so at most $C_2^i$ matches can be played, thus $e_i \ge C_2^{2k} - C_2^i$. Therefore
$$e_{i+1} - b_{i+1} + 1 \ge C_2^{2k} - 2C_2^i = k(2k-1) - i(i-1). \quad (1)$$

When $i > k$, the above can be further improved: consider the first $i$ competitors to arrive and the last $i$ competitors to leave, among which there are at least $2i - 2k$ duplicates. Note that the matches between these duplicated competitors are counted twice in Eq. (1), so when $i > k$, we have
$$e_{i+1} - b_{i+1} + 1 \geq C_2^{2k} - 2C_2^i + C_2^{2i-2k} = (2k-i)^2.$$
Thus the lower bound of the total hotel cost is
$$\sum_{i=0}^{k} (k(2k-1) - i(i-1)) + \sum_{i=k+1}^{2k} (2k-i)^2 = \frac{1}{2}k(4k^2 + k - 1). \quad (2)$$

(2) Construction: Below we provide a construction such that all of the above equalities are satisfied. First divide all competitors into two groups of size $k$, $X = \{X_1, \cdots, X_k\}$ and $Y = \{Y_1, \cdots, Y_k\}$. Next, divide the tournament into the following four segments:
(i) First let $X_1, \cdots, X_k$ check in in this order, and after $X_i$ checks in, have them play against all already-checked-in $X_j$.
(ii) Next let $Y_1, \cdots, Y_k$ check in in this order, and after $Y_j$ checks in, have them play against all $X_i (i > j)$.
(iii) Next let $X_k, \cdots, X_1$ check out in this order, and before $X_i$ checks out, have them play against all $Y_j (i \le j)$.
(iv) Finally let $Y_k, \cdots, Y_1$ check out in this order, and before $Y_i$ checks out, have them play against all remaining $Y_j$.

First, note that for $0 \le s \le k-1$, from the check-in of $T_{k-s}$ to the check-out of $S_{k-s}$, the number of matches during this period is
$$\sum_{j=k-s}^{k-1} (k-j) + 1 + \sum_{j=k-s}^{k-1} (k-j+1) = (s+1)^2. \quad (3)$$

And when $i > k$, $b_{i+1}$ is the arrival day of $T_{i-k+1}$, $e_{i+1}$ is the departure day of $S_{i-k+1}$, thus we have
$e_{i+1} - b_{i+1} + 1 = (2k-i)^2$, which is exactly the minimum value in Eq. (2).

When $i \le k$, note that the $i$ competitors before $b_{i+1}$ all belong to $X$, thus $b_{i+1} = C_2^i + 1$. Similarly, the $i$ competitors after $e_{i+1}$ all belong to $Y$, thus $e_{i+1} = C_2^{2k} - C_2^i$. Thus the equality in Eq. (1) holds. This means that this tournament schedule simultaneously makes both inequalities hold with equality, thus achieving the minimum value.