There is a solution!
Despite my doubts for much of this time, expecting the answer to be that a solution could not be found without going above 150, it turns out that there is a valid solution.
The solution that I have found is:
$$
1,4,6,7,33,50,60,69,94,107,119,127,131,135,142,149
$$
I have confirmed that all 120 sums of two values are unique. Note that a second solution is the same as this, except with each value replaced with 150 minus the value.
The value was found by my code, after maybe 2 days of run time. The code is still running, so it may find more solutions.
Previous observations...
To assist in seeking an answer to this, I have investigated the equivalent questions for smaller numbers. In particular, what's the smallest number of "first $n$ integers" that you need to be able to choose $k$ integers satisfying the condition. For each of the numbers presented here, I have used code to confirm that it cannot be done with a smaller range of values.
In the table below, the set of solutions are "up to reversal" - this is because, if you replace all values, $s$, in the solution with $n+1-s$, you get another solution, and thus all solutions come in equivalent pairs. For example, $[1,2,3,5,8]$ and $[1,4,6,7,8]$ are identical, up to reversal.
$$\begin{array}{c|c|c}
\hline\text{choose $k$ integers} & \text{from first $n$ integers} & \text{Solutions (up to reversal)} \\ \hline
5 & 8 & [1,2,3,5,8] \\ \hline
6 & 13 & \begin{array}{c}[1,2,3,5,8,13]\\ [1,2,3,7,10,13]\end{array} \\ \hline
7 & 19 & \begin{array}{c}[1,2,3,5,9,14,19]\\ [1,2,3,8,11,15,19]\end{array} \\ \hline
8 & 25 & [1,2,3,5,9,15,20,25] \\ \hline
9 & 35 & \begin{array}{c}[1,2,3,5,9,16,25,30,35]\\ [1,2,3,8,14,17,27,31,35]\\ [1,2,3,15,20,25,28,31,35]\end{array} \\ \hline
10 & 46 & [1,2,8,11,14,22,27,42,44,46] \\ \hline
11 & 58 & \begin{array}{c}[1,2,6,10,18,32,35,38,45,56,58]\\ [1,2,6,10,18,32,34,45,52,55,58]\end{array} \\ \hline
12 & 72 & \begin{array}{c}[1,2,3,8,13,23,38,41,55,64,68,72]\\ [1,2,12,19,22,37,42,56,64,68,70,72]\end{array} \\ \hline
13 & 87 & [1,2,12,18,22,35,43,58,61,73,80,85,87] \\ \hline
14 & 106 & \left[\begin{array}{c}1,2,7,15,28,45,55,67,\\70,86,95,102,104,106\end{array}\right] \\ \hline
15 & 127 & \left[\begin{array}{c}1,2,3,23,27,37,44,51,81,\\96,108,111,114,119,127\end{array}\right]\\ \hline
\end{array}$$
Some insights:
The increase in $n$ needed to increase $k$ by 1 seems to grow each time (with the only exception being 13->19->25 - the differences between $n$ values go 5,6,6,10,11,12,14,15,19 - this is not overly surprising, but worth noting.
The density of sums within the possible range can be calculated relatively easily. There are $\frac{k(k-1)}2$ pairs within the set, and thus that many sums if no two are the same. There are $2n-3$ possible sums creatable from a pair of numbers between $1$ and $n$ (as the sum cannot be $1$, $2$, or $2n$). Therefore, the density is given by $\frac{k(k-1)}{2(2n-3)}$ - the resulting densities are (approximately) 76.9%, 65.2%, 60.0%, 59.6%, 53.7%, 50.6%, 48.7%, 46.8%, 45.6%, and 43.5%.
All optimal solutions found so far (accounting for reversal) can be written with $[1,2...$ at the start.
Interestingly, while a solution for $k=15$ can be found with $n=127$, no such solution can be found for $n=128$ if you require both 1 and 128 in the set.
Approaches I have used to try to find solutions:
A code that seeks the solution in an "additive" approach - find which values can be added to the existing set you have, and keep expanding until you run out of values, then backtrack and try a new value. With careful coding to ensure it minimises the search space reasonably well, this can find all optimal solutions for up to $k=14$ within a reasonable amount of time, if you input the correct $n$ in (and can similarly be used to confirm that no lower $n$ will work). However, for larger $n$, it becomes a lot more unwieldy.
A greedy code that seeks the solution in a "subtractive" approach - start with all possible values in the same set (so, 1 through 150 for the actual question), then remove numbers based on which ones will reduce the number of "collisions" (sums that appear multiple times) by the most - with some randomness where multiple numbers can achieve the same reduction... and then add numbers back if they don't substantially increase the number of collisions. This can find many solutions quite quickly for "easy" cases, but due to the random aspect, is also pretty slow to find optimal values for larger $n$. A non-random approach is likely to be very, very slow, as you start with a lot of numbers and there are a lot of options for removal.
Take a smaller solution, such as for $k=12$ ($n=72$), then rescale the numbers and try to fill the gaps. For example, multiplying the first $k=12$ solution by 2 gives $[2,4,6,16,26,46,76,82,110,128,136,144]$. From here, we can find that $[31,89,147]$ is capable of being added, for a total of $k=15$... but despite various attempts, I have not succeeded in reaching $k=16$ using this approach.
Start with three numbers $1\le a<b<c$:
$$S_3 = \{a,b,c\}$$Clearly, $a+b$, $b+c$ and $a+c$ are all distinct, so there do not exist any two pairs which have the same sum.
Given $S_k$, find $S_{k+1}$ by the relation:
$$S_{k+1} = S_k \cup \{(x+y-a+1)\}$$
where $x$ and $y$ are the two largest elements in $S_k$ and $a$ is the smallest element.
We prove by induction that no two pairs in $S_n$ have the same sum. The base case for $n=3$ is already proven, and we assume the claim holds for $S_k$.
Notice that all the pairs in $S_{k+1}$ which do not include this new element (or in other words, belong to $S_k$) have their sum at most $x+y$, while all pairs which include the new element have their sum at least $x+y+1$. So, no two pairs in $S_{k+1}$ have the same sum, since no two pairs in $S_k$ have the same sum.
Now, we need to find $a,b,c$ such that the largest element in $S_{10}$ is less than $100$. Clearly, $a=1, b= 2, c= 3$ minimizes the largest element in $S$.
Finding $S_{10}$, we see that it is the Fibonacci Sequence, where $S_n$ contains the first $n+1$ Fibonacci numbers except $0$.
Thus, $\boxed{S= \{1,2,3,5,8,13,21,34,55,89\}}$ is a counterexample.
Best Answer
A picture is worth a thousand words:
... but I will write a few words, nevertheless. This configuration has many symmetries and interesting properties, some of which I will describe here; however, I recommend trying to explore the configuration and figure them out on your own before reading further.
The eight points are arranged in two squares of different side lengths, with the same centre, and at a 45 degree angle from each other, the internal square coloured red and the external coloured blue. There are also eight equilateral triangles with vertices from the eight points: each side of the internal red square is completed to an outward-pointing equilateral red triangle, and each side of the external blue square is completed to an inward-pointing blue triangle.
All twelve red segments have equal lengths, as do all twelve blue segments. These equalities suffice to verify that any pair of points has a pair of two equidistant points. Up to symmetries, there are only six cases to check: two where both vertices are from the blue square (adjacent or opposite); two where both vertices are from the red square (adjacent or opposite); and two where each vertex is from a different square (and the distance between the two vertices is either red or blue). The final two cases are the nicest and most interesting, and the latter of which is shown in the diagram, where the green perpendicular bisector of the two green points passes through a different pair of points, obtained from the first pair by a 90 degree rotation.
I do not know the origin of the problem, but I have encountered it before—on the xkcd forums (or "fora"), of all places, at least 10, maybe 15 years ago. I had recently recalled this problem and had presented it to some of our IMO students. My attempt to look up the original formulation and my original answer led me here. Sadly, the xkcd fora seem to have been offline since August 2019, and apparently they were not archived in full (or at least, I am unable to find such an archive), and thus some small pages in the history of this question might now be forever lost...