A finite set of distinct positive numbers is special if each integer in the set divides the sum of all integers within the set.
Here is a partial answer. Clearly, it suffices to show that $[n]=\lbrace 1,2,3,\ldots,n \rbrace$ is contained in a special set for every $n$, since any finite set of positive integers is included in some $[n]$. I describe an algorithm below that I have checked to work on every $[n]$ for $8 \leq n \leq 20$.
Here is the algorithm. It starts with an initial finite set $A$ of positive integers, which we increase one element at a time, until we hit a special set.
Step 1. Compute the sum $s=\sum_{a\in A} a$.
Step 2. Compute $X_1=\lbrace a \in A \ | \ a\not\mid s \rbrace$. If $X_1$ is empty, then $A$ is special and we are done. Otherwise, let $x_1$ be the smallest element in $X_1$.
Step 3. Compute $X_2=\lbrace a \in A \ | \ a\mid s \rbrace$ (so $X_2$ is the complement of $X_1$ in $A$). Denote by $l$ the lcm of the elements of $A$ (in particular, $l=1$ if $X_2$ is empty).
Step 4. Let $M$ be the smallest integer that satisfies the following three conditions : (1) it is larger than the largest element of $A$, (2) it is divisible by $l$, (3) the sum $s+M$ is divisible by $x_1$ (note that the congruence conditions are compatible by construction).
Step 5. Replace $A$ with $A'=A\cup \lbrace M \rbrace$ and return to step 1.
When $n=50$ for example, the algorithm eventually produces the 99-element special set
$$ [50]\cup\lbrace 1275, 2550, 30600, 35700, 142800, 2142000, 28274400, 30630600, 1102701600, 25607181600, 53542288800, 2248776129600, 69872686884000, 72201776446800, 5198527904169600, 213717258282528000, 9200527969062830400, 433301055304911393600, 2656323860782282891200, 12396178016983986825600, 30990445042459967064000, 464856675636899505960000, 511342343200589456556000, 5113423432005894565560000, 6136108118407073478672000, 269988757209911233061568000, 1129043893786901520075648000, 29637402211906164901985760000, 31048707079139791802080320000, 1241948283165591672083212800000, 24776868249153553858060095360000, 469456451036593652047454438400000, 8424135204712208311740432422400000, 142714761115124470222426149273600000, 2274516505272296244169916754048000000, 33966113145399623912937423527116800000, 473099433096637618787342684841984000000, 6113900366171932304328736234881024000000, 72857312696882193293250773465665536000000, 794807047602351199562735710534533120000000, 7868589771263276875671083534291877888000000, 69943020189006905561520742527038914560000000, 550801283988429381296975847400431452160000000, 3776923090206372900322120096460101386240000000, 22032051359537175251879033896017258086400000000, 105753846525778441209019362700882838814720000000, 396576924471669154533822610128310645555200000000, 1057538465257784412090193627008828388147200000000, 1586307697886676618135290440513242582220800000000\rbrace $$
Say we're given a set $S$, with sum $s$. We assume that $S$ does not consist of only powers of $2$; if it does, we can simply add to the set the number $3$. First, let $a$ be large enough so that $2^a > 2s$, meaning $2^a - s \not \in S$, and define $S' = S \cup \{2^a - s\}$, so $S'$ has sum $2^a$. Let $n$ be the product of all elements of $S'$, and let $b$ be large enough so that $2^b > n$.
We now construct a set $S''$ containing $S'$ with sum $2^{a+b} n$, all elements of which divide $2^{a+b} n$. Since $n-1$ is less than $2^b$, using its binary representation we can express $n-1$ as a sum of distinct elements of $\{1, 2, 4, \dots, 2^{b-1}\}$, and thus we can express $2^a(n-1)$ as a sum of distinct elements of $\{2^a, 2^{a+1}, \dots, 2^{a+b-1}\}$. Let $T$ be the subset of elements appearing in the latter sum. Then define $$S'' = S' \cup T \cup \{2^an, 2^{a+1}n, \dots, 2^{a+b-1}n\}.$$ As you can check, all elements of $S''$ divide $2^{a+b} n$, and the three sets in this union are disjoint (since $n$ is not a power of $2$), and thus $S''$ has sum $2^a + 2^a(n-1) + (2^{a+b} n - 2^a n) = 2^{a+b} n$, meaning $S''$ is special.
TL;DR: With any non-special set $A$ of distinct integers, let the sum of the elements be $s$. The $\operatorname{lcm}$ of all of the elements of $A$, call it $q$, can always become a practical number, call it $m$, by multiplying with an appropriate integer. There then exists a set $B$ of distinct multiples of $s$, where each element divided by $s$ is a factor of $m$ and the sum of the elements is $(m - 1)s$. Then $A \cup B$ is a special set.
If the finite set of positive integers is a special set itself, then you can use just it. In particular, any single integer itself forms a special set, so if $n$ is the number of elements, any non-special set has $n \gt 1$. Also, in those cases, have the set be $A = \{a_i\}_{i=1}^{n}$ and let
$$s = \sum_{i=1}^{n}a_i \tag{1}\label{eq1A}$$
Consider adding multiples of $s$ to form a special set. For example, if $A = \{2,3\}$, then $s = 5$, with $2(5) = 10$ and $3(5) = 15$ being sufficient to add on to form a special set with a new sum of $30 = (2)(3)5$. In general, at the minimum, the new total sum must have a factor of the $\operatorname{lcm}$, call it $q$, of all of the $a_i$, plus $s$ must divide the sum as well, but it can have more factors than this if need be.
For some $j \ge 1$, let $B = \{b_i(s)\}_{i=1}^{j}$, where $b_i$ are distinct positive integers, be a set of multiples of $s$ which are being added, to get
$$S_t = s + \sum_{i=1}^{j}b_i(s) = s(1 + \sum_{i=1}^{j}b_i) \tag{2}\label{eq2A}$$
where $S_t$ is the total sum of the elements in $A \cup B$. Next, let
$$m = 1 + \sum_{i=1}^{j}b_i \tag{3}\label{eq3A}$$
You must have $b_i \mid m \; \forall \; 1 \le i \le j$, plus $q \mid ms$.
Note a practical number is
... a positive integer $n$ such that all smaller positive integers can be represented as sums of distinct divisors of $n$.
This means if $m$ is a practical number, there are distinct $b_i$, which are all factors of $m$, that give $\sum_{i=1}^{j}b_i = m - 1$. Regarding the requirements to be a practical number, the Characterization of practical numbers section states
A positive integer greater than one with prime factorization $n=p_1^{\alpha_1}\ldots p_k^{\alpha_k}$ (with the primes in sorted order $p_1 \lt p_2 \lt \dots \lt p_k$) is practical if and only if each of its prime factors $p_{i}$ is small enough for $p_{i}-1$ to have a representation as a sum of smaller divisors. For this to be true, the first prime $p_{1}$ must equal $2$ and, for every $i$ from $2$ to $k$, each successive prime $p_{i}$ must obey the inequality $$p_{i} \leq 1 + \sigma(p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{i-1}^{\alpha_{i-1}}) = 1 + \prod_{j=1}^{i-1}\frac{p_{j}^{\alpha_{j} + 1} - 1}{p_{j} - 1}$$ where $\sigma(x)$ denotes the sum of the divisors of $x$.
As previously stated, you can add more factors if needed, e.g., just a sufficiently large power of $2$ or, alternatively, any one or more single or multiple factors of any primes up to the largest required prime. In any case, this means you can always easily create an $m$ which is a practical number and which satisfies the other conditions, resulting in $A \cup B$ forming a special set.