, 2022

The condition of the problem requires $\binom{n}{k}$ to be divisible by $n$ for all $1 \le k \le \sqrt{n}-1$. Assume there exists a prime divisor $p$ of $n$ such that $p \le \sqrt{n}-1$ and set $k = p$. Since
$$p! \binom{n}{p} = n(n-1)(n-2)\cdots(n-p+1)$$
and $\binom{n}{p}$ is divisible by $n$, we have that $p$ is a divisor of $(n-1)(n-2)\cdots(n-p+1)$. This assertion is not true since $p$ is a factor of $n$ and thus it is not a divisor of any of $n-\ell$, $1 \le \ell \le p-1$. From now on we assume that all prime factors of $n$ are greater than $\sqrt{n}-1$. Assume $n$ has at least three prime factors (counting $p^\alpha$ as $\alpha$). Direct verification shows that $n=8$ is a solution and for $n=12$ we have factor $2 < \sqrt{12}-1$. For $n \ge 16$ we obtain
$n > (\sqrt{n}-1)^3 \ge (\frac{3\sqrt{n}}{4})^3 = \frac{27n^{3/2}}{64}$, i.e. $n < (\frac{64}{27})^2 < 9$, a contradiction. It remains to consider $n = pq$ for primes $p$ and $q$ such that $\sqrt{n}-1 < p \le q$. If $q \ge p+3$ then $n \ge p(p+3) > (\sqrt{n}-1)(\sqrt{n}+2) = n + \sqrt{n}-2 > n$, a contradiction. If $q = p+1$ we have $p=2, q=3$, i.e. $n=6$ which is a solution. If $q = p$ and $1 \le k \le p-1$ then the left hand side of
$$k! \binom{p^2}{k} = (p^2 - k + 1)(p^2 - k + 2) \cdots (p^2 - 1)p^2$$
is divisible by $p^2$ and $k!$ is not divisible by $p$. Therefore $p^2$ is a factor of $\binom{p^2}{k}$. If $q = p+2$ and $1 \le k \le p-1$ then the left hand side of
$$k! \binom{p^2+2p}{k} = (p^2 + 2p - k + 1)(p^2 + 2p - k + 2) \cdots (p^2 + 2p - 1)(p^2 + 2p)$$
is divisible by both $p$ and $p+2$. Since $k!$ is not divisible by both $p$ and $q = p+2$ we infer that $\binom{p^2+2p}{k}$ is divisible by $p$ and $p+2$, i.e. it is divisible by $p(p+2)$. Since $2022 < 2025 = 45^2$ all primes $2 \le p < 45$ imply the solution $n = p^2$. The primes for which $p+2$ is also a prime imply also the solution $n = p(p+2)$. Direct verification shows that there exist 14 primes in this interval: $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43\}$, and for 6 of them ($\{3, 5, 11, 17, 29, 41\}$) $p+2$ is also a prime. Finally, the desired number is (adding 6 and 8): $14+6+2=22$.