This is not a solution, just something I found that might be relevant.
On the page linked to in the question, a reduction and various strategies are considered. I'll briefly reproduce the reduction, both because I think it's the most useful part of that page and perhaps not everyone will want to read that entire page, and also because I need it to say what I found.
Let a counterexample with minimal $n$ be given. If one of the sticks were of length $n$, we could use that stick as the target stick of length $n$ and cut the remaining sticks into lengths $1$ through $n-1$, since otherwise they would form a smaller counterexample. Likewise, if one of the sticks had length greater than $2n-2$, we could cut off a stick of length $n$ and the remaining sticks would all be of length $\ge n-1$, so again we could cut them into lengths $1$ through $n-1$ because otherwise they would form a smaller counterexample. Thus,
the lengths of the sticks in a counterexample with minimal $n$ must be $\gt n$ and $\lt 2n-1$.
Problem instances that satisfy these conditions for a potential minimal counterexample are called "hard" on that page; I suggest we adopt that terminology here.
The strategies discussed on that page include various ways of forming the target sticks in order of decreasing length. It was found that there are counterexamples both for the strategy of always cutting the next-longest target stick from the shortest possible remaining stick (counterexample $\langle11,12,16,16\rangle$) and for the strategy of always cutting the next-longest target stick from the longest remaining stick unless it already exists (counterexample $\langle10,10,12,13\rangle$), whereas if the stick to cut from was randomized, it was always possible to form the desired sticks up to $n=23$.
I've checked that all hard problem instances up to $n=30$ are solvable, and I found that they remain solvable independent of which stick we cut the target stick of length $n$ from. This is equivalent to saying that a problem instance for $n-1$ can always be solved if all stick lengths except one are $\gt n$ and $\lt 2n-1$ and one is $\lt n-1$, since all of these
instances can result from cutting a stick of length $n$ from a hard problem instance for $n$.
I thought that this might be generalized to the solvability of an instance being entirely determined by whether the sticks of length $\le n$ can be cut to form distinct integers, but that's not the case, since it's possible to leave only a few holes below $n$ such that the few remaining sticks above $n$ can't fill them.
We answer the following seemingly more general question. Find all triangles such that the sides of the triangle and one of the altitudes are four integers which are, in some order, consecutive terms of an arithmetic progression. We can assume that these four terms are $q+d$, $q$, $q-d$, and $q-2d$, where $d$ is positive and $q$ and $d$ are relatively prime.
We will use Heron's Formula for the area of a triangle in terms of its sides. If a triangle has sides $a$, $b$, and $c$, then the area of the triangle is
$$\sqrt{s(s-a)(s-b)(s-c)},\quad \text{where} \quad s=\frac{a+b+c}{2}.$$
At least two of the sides of the triangle are greater than the altitude. There are two sorts of cases to consider: (1) The altitude is the second smallest of the numbers $q+d$, $q$, $q-d$, and $q-2d$ and (2) The altitude is the smallest of these four numbers.
Case 1: We have $s=(3q-d)/2$. Thus by Heron's Formula the area of the triangle is $\sqrt{(3q-d))(q-3d)(q-d)(q+3d)/16}$. But this area is also $(1/2)(q-2d)(q-d)$. Squaring, we obtain
$$\frac{(3q-d)(q-3d)(q-d)(q+3d)}{16}=\frac{(q-2d)^2(q-d)^2}{4}.$$
Cancel a $q-d$ from both sides, multiply through by $16$, and simplify. After a while we arrive at
$$q^3-19q^2d+59qd^2-25d^3=0.$$
We look for solutions of this equation in relatively prime positive integers. Since $q$ divides the first three terms of the equation, it must divide $25d^3$. Since $q$ is relatively prime to $d$, $q$ must divide $25$, which leaves the possibilities $q=1$, $q=5$, and $q=25$. None of these works.
Case 2: Heron's Formula shows that the triangle has area $\sqrt{(3q)(q-2d)(q)(q+2d)/16}$. There are unfortunately three possibilities: (i) The altitude is to the side $q$; (ii) The altitude is to the side $q+d$; (iii) The altitude is to the side $q-d$.
In each case, by Herons's Formula, the area of the triangle is $\sqrt{(3q)(q-2d)(q)(q+2d)/16}$.
(i): The area is then $(q)(q-2d)/2$. We obtain the equation
$$\frac{(3q)(q-2d)(q)(q+2d)}{16}=\frac{(q)^2(q-2d)^2}{4}.$$
Divide through by $q^2(q-2d)$, multiply by $16$, and simplify. We obtain $q=14d$, so since $q$ and $d$ are relatively prime we have $q=14$ and $d=1$. That gives us the triangle with sides $15$, $14$, $13$, and height $12$ (together with all integers scalings of this triangle).
(ii): The area is $(q+d)(q-2d)/2$. We obtain the equation
$$\frac{(3q)(q-2d)(q)(q+2d)}{16}=\frac{(q+d)^2(q-2d)^2}{4}.$$
This simplifies to $q^3-6q^2d -12qd^2-8d^3=0$. We look for relatively prime positive integer solutions. Since $q$ divides the first three terms, it must divide $8$, and it is easy to check that none of $1$, $2$, $4$, or $8$ work. (Of course $1$ and $2$ are irrelevant, since they would not give a triangle).
(iii): The area is $(q-d)(q-2d)/2$. We obtain the equation
$\frac{(3q)(q-2d)(q)(q+2d)}{16}=\frac{(q-d)^2(q-2d)^2}{4}.$ This simplifies to $q^3-22q^2d +20qd^2-8d^3=0$, which again has no solutions in relatively prime integers.
Comments: In Case 1, there are up to similarity two solutions in which our four numbers form an arithmetic progression, and in each of Cases 2(ii) and 2(iii) there is one, but for all of these $q$ and $d$ are incommensurable.
The original problem in principle does not require much machinery. If $q$ is not small, then the numbers from $q-2$ to $q+1$ are (relatively) close to each other, so the triangle looks more or less equilateral. But an altitude of an equilateral triangle is not close to a side in length. So we can see that large $q$ are not possible, and it comes down to checking a smallish number of cases.
However, the algebra is more efficient.
Best Answer
The lengths of the sticks can be arranged in the nondecreasing series $(a_1,a_2,...,a_{12})$. The shortest length is $1$, so we can say, that in this bag there are two sticks with length $1$, so $a_1=a_2=1$
Now we can construct next pieces by taking two longest (last) sticks, summing their lengths and adding $1$ to obtain the smallest integer number larger than the sum of two other integers: $$a_{k+2}=a_k+a_{k+1}+1$$ we have then: $$a_{k+2}>a_k+a_{k+1} \geq a_i+a_j \,\,\,(i<j\leq k+1) $$ so if we get $3$ sticks, the longest is longer than the sum of the remaining two, so we can't arrange them to make a triangle.
The series is then $$(1,1,3,5,9,15,25,41,67,109,177,287)$$
We can also say, that the degenerate triangles doesn't count as triangles in this context. In this case we don't have to add $1$ to the lengths of two sticks in forst formula, so we have: $$a_{k+2}=a_k+a_{k+1}$$ and this is the recursive formula for Fibonacci series: $$(1,1,2,3,5,8,13,21,34,55,89,144)$$