Solution:

I hate constructions.

The answer is $k=3$, which we can make with $A_1=\{1,2,6,10,13,16,19,22,\ldots\}$, $A_2=\{7,8,9,11,14,17,20,23,\ldots\}$, and $A_3=\{3,4,5,12,15,18,21,24,\ldots\}$. To show that $k>3$ does not work, we focus on partitioning $1,2,\ldots,19$ into the sets and only considering $n\le20$. Each set in the partition must achieve the sums $15,16,17,18,19,20$, so we must have that $|A_i|\ge4$ for all $i$, ruling out $k>4$. Furthermore, the only possible set $\{a,b,c,d\}$ of four elements with the set of sums of distinct pairs of elements equal to $\{15,16,17,18,19,20\}$ is $\{7,8,9,11\}$. Indeed, WLOG $a<b<c<d$, and note that we must have $a+b=15$ and $c+d=20$. This implies that $b\ge8$ and $c\le9$ as $c\ne d$, so as $b<c$ we must have $(a,b,c,d)=(7,8,9,11)$. Now if $k=4$, then we must have three sets of five and one set of four, which must be $\{7,8,9,11\}$. We claim that the pairs $(1,19),(2,18),(3,17),(4,16),(5,15)$ must each be in different sets of the partition, which contradicts $k=4$. We proceed with strong induction. For the base case, consider the set that $19$ is in. If $19$ does not contribute to the pairwise sums among $15,16,17,18,19,20$, then the remaining four numbers must pairwise add up to $15,16,17,18,19,20$, meaning that they are $7,8,9,11$, a contradiction. Hence, $19$ must contribute, meaning that $1$ is in the same set as $19$. For the inductive step, suppose that the pairs $(1,19),(2,18),\ldots,(k-1,21-k)$ are together in different sets of the partition. We will show that $k$ and $20-k$ must be together in a different set of the partition. Consider the set that $20-k$ is in. If it is in the same set as $a$ and $20-a$, then note that $20-k$ only contributes the sum $20-k+a$ to the set. Indeed, $20$ is already a sum of the set, and among $20-k+1,20-k+2,\ldots,19$, it can only contribute $20-k+a$ to the set as all numbers in $1,2,\ldots,k-1$ other than $a$ are already in different sets of the partition. But as $20-a$ also cannot contribute more to the set of sums than just $20$ (as $1,2,\ldots,a-1$ are in different sets of the partition), we must have that $a$ and the other two elements in the set contribute four sums altogether, a contradiction. Hence, $20-k$ is not in any of the previous sets. But $k$ must also be in the same set as $20-k$, as otherwise the rest of the four numbers would again need to be $7,8,9,11$ for the set to achieve all six sums. This completes the induction and produces our contradiction.

Construction for $k=3$ is $$\{1, 2, 3, 3k+2|k\ge 4\}$$$$\{4, 5, 6, 10, 3k+1|k\ge 4\}$$$$\{7, 8, 9, 11, 3k|k\ge 4\}$$ To prove $k>3$ doesn't work, note that for all $i$, $A_i$ must have at least 5 terms less than or equal to 20. This is because all numbers less than or equal to 21 can obviously only be expressed as a sum of two positive integers less than or equal to 20, and we have 7 sums we need to make less than or equal to 21: 15, 16, 17, 18, 19, 20, and 21. Since $\binom{4}{2}=6$, each $A_i$ needs 5 or more terms to form all of the desired sums. This automatically means we can have at most $k=4$. Now we prove $k=4$ doesn't work. Assume for the sake of contradiction that it does. Assume without loss of generality that $A_1$ contains $1$. Now consider the three sets $\{1, 2, 3, 4, 5, 6, 7\}, \{8, 9, 10, 11, 12, 13\}, \{14, 15, 16, 17, 18, 19, 20\}$ If $A_1$ has four elements in the third set, then obviously $\binom{4}{2}=6$ of those sums are too big so we can have max 4 valid pairs. If $A_1$ has three elements in the third set, $\binom{3}{2}$ of those sums are too big, and the last element+1 is too small, so we can have max $10-4=6$ valid pairs. If $A_1$ has two elements in the third set, it cannot have any elements in the second set, because the second set+1 numbers are two small and the second set+third set numbers are too big for our range. This means it must have three elements in the first set and two in the third set. Thus we have at least 4 pairs that don't work, so we have max 6 valid pairs. If $A_1$ has one element in the third set, then it must have three in the second set, because 1+anything in the first or second sets is too low, and any two in the first set is too low. However, having three in the second set+the one in the third set makes three more invalid sums. If $A_1$ has no elements in the third set, then clearly 1+other elements give 4 sums that are too low. In all cases, we cannot reach 7 valid sums, so thus we have a contradiction.

The answer is $k = 3$. To show that $k \geq 4$ does not work, consider truncating the sets $A_1, A_2, \dots, A_k$ to $B_1, B_2, \dots, B_k$ such that none of their elements are greater than 18. It follows that each of these sets contains two distinct elements whose sum is $n$ for $15 \leq n \leq 19$. There exist at least two $B_i$ which have at most 4 elements. We claim that this is possible. This is immediate from the following claim. Claim. If $S$ is a set of 4 positive integers such that for each $15 \leq n \leq 19$, there exist distinct elements of $S$ whose sum is $n$, then $9 \in S$. Proof. Suppose $S = \{a,b,c,d \}$ with $a \leq b \leq c \leq d$. Since $\binom{4}{2} = 6$, it follows that every pair of elements must sum to an element of $\{15, 16, 17, 18, 19\}$, except one, which we call the weird pair. Suppose that $(a,b)$ is the weird pair, implying $a \leq 6$. Therefore $c \geq 9$, since $a + c = 15$. However, we have $c + d = 19$ since it is the maximal pair, forcing $c=9$ and $d = 10$. Of course, this also forces $a = 6$. It is easy to conclude from here that $b = 8$. Hence $S = \{6,8,9,10\}$ is the only possibility here. Suppose that $(c,d)$ is the weird pair, implying $d \geq 11$. This is case is symmetric to the previous. Therefore we have $b \leq 8$, since $b + d \leq 19$. We also have $a + b = 15$ since it is minimal, forcing $b=8$ and $a =7$ and $d = 11$. Therefore $c = 9$. This gives $S = \{7,8,9,11\}$. If neither $(a,b)$ nor $(c,d)$ are the weird pair, then $a + b =15$ and $c + d = 19$. This is only possible if $S = \{7, 8, 9, 10\}$. This establishes the claim. $\blacksquare$ It follows that $9$ belongs to two separate sets $B_i$ and $B_j$, which is is impossible. Therefore $k < 4$. Consider the following construction for $k = 3$. \begin{align*} A_1 &= \{1,2,3\} \cup \{ 3k + 3 \mid k \geq 3 \} \\ A_2 &= \{4,5,6\} \cup \{3k+1 \mid k \geq 3 \} \\ A_3 &= \{7,8,9\} \cup \{ 3k+2 \mid k \geq 3 \} \end{align*}Then each integer $n$ can be made the sum of one element on the right set and one element on the left set.

Solution: we can only write $15$ as a sum of 2 distinct positive integers if both numbers are in $[1,14],$ so each $A_i$ must have at least 2 numbers in $[1,14]$ implying $k\leq 7.$ Notice that a set must contain at least 2 numbers in $[1,14]$, if both are greater than $1$, since either $a+1$ or $16-a$ is also in, it's a contradiction if there are exactly 2. so $k\leq 5,$ and $k=5$ is the case that the set that contains $1,14$ doesn't contain $2,3,...13$ so it must contain $15,16,...,28,$ obviously impossible. Thus $k\leq 4$ then if there are exactly 3 elements in a set in $[1,14]$, namely, $b,15-b,b+1$with $b\in [3,12]$ then with similar argument, contradiction, observe that there can be at most 1 set with exactly 3, so $14\geq 3+4+4+4,$ contradiction.Hence $k\leq 3,$ and $3$ is constructable as above posts.

The answer is \(k=3\), achieved by \begin{align*} A_1&=\{1,2,3\}\sqcup\{12,15,18,\ldots\}\\ A_2&=\{4,5,6\}\sqcup\{11,14,17,\ldots\}\\ A_3&=\{7,8,9\}\sqcup\{10,13,16,\ldots\}. \end{align*}Now we will show \(k\ge4\) fail. Say a set \(A\) is encompassing if for each \(n\ge15\), there are two distinct elements of \(A\) whose sum is \(n\). If \(k\ge4\), for some \(i\), \[|A_i\cap\{1,2,\ldots,23\}|\le5\]by Pigeonhole. I contend, then, that \(A_i\) cannot be encompassing. In general, we will show that the set \[S=\{a,b,c,d,e,24,25,26,\ldots\}\]is not encompassing whenever \(a<b<c<d<e<24\). Evidently this will suffice. With the exception of the \(\binom52=10\) sums \(a+b\), \(a+c\), \ldots, \(d+e\), the remaining sums of distinct elements of \(S\) are all at least \(24+a\). Therefore the 10 sums must cover all numbers among 15, 16, \ldots, \(23+a\). In particular, \(a=1\), and \[\{a+b,a+c,\ldots,d+e\}=\{15,16,\ldots,24\}.\]The two smallest elements of \(\{a+b,a+c,\ldots,d+e\}\) are \(a+b\) and \(a+c\), so \(a+b=15\) and \(a+c=16\). This implies \(b=14\) and \(c=15\), hence \(b+c=29>24\), contradiction. Remark: The choice of 23 minimizes the amount of work that needs to be done. The choices of 19 and 15 also work, albeit they require some more effort and require considering the other \(k-1\) sets. For instance, here is a sketch by using 15 instead of 23. We can show the following, which suffice: when \(|A_i\cap\{1,2,\ldots,15\}|\le3\), we have \(A_i\cap\{1,2,\ldots,15\}=\{1,2,14\}\); when \(|A_i\cap\{1,2,\ldots,15\}|\le4\), we have \(\min A_i\le4\).

The answer is $3$, with the construction \begin{align*} 1,2,3,14,17,20,23,\dots \\ 4,5,6,10,13,16,19,\dots \\ 7,8,9,11,12,15,18,\dots \end{align*}Suppose $k\ge 4$. Then, we will focus on partitioning only the first $19$ positive integers to satisfy that all numbers between $15$ and $20$ inclusive appear as sums. There are six sums, so each set must have at least $4$ numbers. Thus, $k=4$. Even so, there must be a set with exactly four numbers. Let the numbers be $a<b<c<d$. Then $a+b=15$, $c+d=20$, $a+c=16$, and $b+d=19$. As such we have $d=a+4$ and $c=b+1$. Additionally, $b+c$ and $a+d$ are $17$ and $18$ in some order, and $b+c=2b+1$ is odd, so $2b+1=17$, $b=8$. We similarly get $a=7$. Thus, our first set must be $\{7,8,9,11\}$. Now, note that no other set may have just four elements, and so each other set must have exactly five. Resultingly, no element can be "irrelevant." That is, for each element $x\in A_i$, there exists $y\in A_i$ such that $15\le x+y\le 20$. For example, $1$ must be in the same set as $19$ because otherwise removing $19$ will not remove any sums, meaning that there's another $4$ element set that works, impossible. Let $A_1=\{7,8,9,11\}$ and $\{1,19\}\subset A_2$. Note that of $A_2\setminus 19$ there is every sum except $20$. Since there are four elements, there is at most one pair of elements $i,j\in A_2\setminus 19$ such that $i+j$ is not between $15$ and $19$ inclusive. If $1$ is not one of those elements, then the other three elements must all be at least $14$, which implies any pairwise sum between them is at least $28$, absurd. Thus, let $i=1$. Then, let the other three be $a_1<a_2<a_3$. We have $a_2+a_3<20$ so $a_2\le 9$. Hence, $1+a_1$ and $1+a_2$ are both not between $15$ and $19$ inclusive. Contradiction.

Solved with mananaban. The answer is $\boxed{3}$. The construction is \[\color{red}{1\;2\;3}\; \color{blue}{4\;5\;6}\; \color{green}{7\;8\;9}\; \color{green}{10}\; \color{blue}{11}\; \color{red}{12}\; \color{green}{13}\; \color{blue}{14}\; \color{red}{15}\; \color{green}{16}\; \color{blue}{17}\; \color{red}{18}\;\dots \]which clearly works. Now we will prove that $k=4$ doesn't work. This clearly finishes. Consider forming $15$ to $21$. Since there are $7$ numbers, we need at least $5$ numbers in each set that are at most $20$. Thus each set has exactly $5$ numbers below $21$. Consider the set containing $20$. Then the $20$ can only contribute to covering $21$, meaning that the remaining four numbers must cover $15$ to $20$(thus they cannot also cover $21$). However, this means that $1$ is in the set. Then the set also has $a,b$ such that $1+a,1+b\in\{15,16,\dots,20\}$, impossible as then $a+b\ge 29$. $\blacksquare$