I am writing a long comment on this question because it is related to a fascinating mystery about "middle-level designs" that I have been wondering about for several years.
As written in the (now deleted AI-generated) other answer, what you are asking for is exactly a (self-complementary) Steiner $S(k, n, 2n)$ system. The well-known divisibility conditions for designs requires in this case that $$\binom{n - i}{k - i} \mid \binom{2n-i}{k-i} \qquad (0 \le i \le k).$$ Moreover, $k \ge n/2$ by the pigeonhole argument of Matthew Bolan. This is a curious case in which we can (almost) exactly solve the divisibility conditions.
Observation: If $k = n-1$ then the divisibility conditions are satisfied if and only if $n+1$ is prime. If $k < n-1$ then the divisibility conditions are probably never satisfied (i.e., I haven't proved it).
In particular, OP's example $n=11$ is not "$k$-splittable" for any $k < n$, since $n+1 = 12$ is not prime.
Indeed, let $s = n - k \le n/2$ and $j = n - i$. Then the divisibility condition is $$\frac{j!}{s!} \mid \frac{(n+j)!}{(n+s)!} = (n+s+1) \cdots (n+j) \qquad (s < j \le n).$$ If $s = 1$ and $n+1 = p$ is prime then it reduces to checking that $p$ divides $\binom{n+j}{j}$ for all $j = 1, \dots, n$, which is clear.
In the other direction, suppose $p$ is a prime factor of $n+i$ where $1 \le i \le s$. If $p \ne n+i$ then $p \le (n+i)/2 < n$. If moreover $p > s$ then the divisibility condition gives $$p \mid \frac{p!}{s!} \mid (n+s+1) \cdots (n+p),$$ which is impossible because $p$ divides $n+i < n+s+1$ and $n+i+p > n+p$. It follows that every integer in the interval $[n+1, n+s]$ is either prime or $s$-smooth. This is extremely rare for $s > 1$ and $n \ge 2s$. For $n > 15$ the only solutions I know have $s \in \{2,3\}$. For $2 \le s \le 5$, here are some case-specific arguments:
Case $s = 2$: We have $4 \mid 4!/2! \mid (n+3)(n+4)$, so $(n+1)(n+2)$ is not divisible by $4$. Since either $n+1$ or $n+2$ is $2$-smooth we must have $n+1 = 2$ or $n+2 = 2$, a contradiction since $n \ge 2s = 4$.
Case $s = 3$: We have $4 \mid 4!/3! \mid n+4$, so $n$ is divisible by $4$ and $n+2 \equiv 2 \pmod 4$. Since $n+2$ is either prime or $3$-smooth it follows that $n+2 = 2 \cdot 3^k$. Now the divisibility condition with $j=9$ and considering just the factors of $3$ gives $$3 \cdot 9 \mid (n+5)(n+8) = (2 \cdot 3^k + 3)(2 \cdot 3^k + 6),$$ which implies $k \le 1$, so $n \le 4$, a contradiction since $n \ge 2s = 6$.
Case $s = 4$: Two of $n+1, n+2, n+3, n+4$ must be prime and the other two must have the form $2\cdot 3^a$ and $2^b$, which implies that $|3^a - 2^{b-1}| = 1$. It follows that $n = 15$.
Case $s = 5$: Since $6 = 6! / 5! \mid n+6$, we have $6 \mid n$ and therefore $2 \mid n+2, n+4$ and $3 \mid n+3$. Since each of $n+1,n+2,n+3,n+4,n+5$ is supposed to be either prime or $5$-smooth, it must be that one of $n+2, n+4$ is a power of $2$, the other has the form $2 \cdot 5^k$, and $n+3$ is a power of $3$, so again the configuration contains a power of $2$ and a power of $3$ differing by $1$, so $n = 6$, a contradiction since $n \ge 2s = 10$.
One further comment on this strategy: If every integer in the interval $[n+1, n+s]$ is prime or $s$-smooth then by restricting to the even integers and dividing by $2$ we have an interval of at least $(s-1)/2$ integers $> s$ that are all $s$-smooth. This kind of problem has a long history in number theory, and for example Erdős (see Section 7 of [HT] -- thank you to Khalid Younis for this reference) proved that the longest interval of $s$-smooth integers greater than $s$ has length $O(s / \log s)$. Therefore we are done for sufficiently large $s$, and there might even be a bound in the literature somewhere that gives us a complete solution -- I haven't checked sufficiently.
Finally, let us go back to the case $s = 1$, i.e., $k = n-1$. As explained above, the divisibility conditions are satisfied if and only if $n+1 = p$ is prime. But, the divisibility conditions do not guarantee the existence of a design, they are merely necessary conditions. Thus we face the following mysterious question.
Question: Is there a Steiner $S(n-1, n, 2n)$ system whenever $n+1 = p$ is prime?
The answer is yes for $p = 2, 3$ (trivial cases), $p = 5$ ($3$-dim affine geometry over $\mathbf{F}_2$), and $p = 7$ (Witt design on 12 points, whose automorphism group is $M_{12}$). For $p \ge 11$ I believe the question is open.
As for self-complementarity, a result of Mendelsohn [M] implies that every $S(n-1, n, 2n)$ system is automatically self-complementary. Moreover, there is a $S(n-1, n, 2n)$ system if and only if there is a $S(n-2, n-1, 2n-1)$ system.
[HT] Hildebrand, Adolf; Tenenbaum, Gérald, Integers without large prime factors, J. Théor. Nombres Bordx. 5, No. 2, 411-484 (1993). ZBL0797.11070.
[M] Mendelsohn, N. S., A theorem on Steiner systems, Can. J. Math. 22, 1010-1015 (1970). ZBL0192.33305.
