Let $$X$$ be the set consisting of twenty positive integers $$n, n+2,...,n+38$$. The smallest value of $$n$$ for which any three numbers $$a, b, c \in X$$, not necessarily distinct, form the sides of an acute-angled triangle is:

Let the set be $$X=\{n,n+2,n+4,\dots ,n+38\}$$. These are 20 consecutive even (or odd) integers with common difference $$2$$.

Pick any three elements of $$X$$ and arrange them in non-decreasing order $$a\le b\le c$$. They must satisfy two conditions:

1. Triangle inequality $$a+b\gt c$$.
2. Acute-angled condition $$a^2+b^2\gt c^2$$.

The left-hand sides of both inequalities increase when we replace $$a$$ or $$b$$ by larger elements of $$X$$, while the right-hand side decreases if we replace $$c$$ by a smaller element. Therefore the “hardest” (most restrictive) triple is obtained by taking the two smallest and the largest numbers from the set:

$$a=b=n,\qquad c=n+38$$.

Checking the triangle inequality for this extreme triple:

$$n+n \gt n+38\; \Longrightarrow\; 2n\gt n+38\; \Longrightarrow\; n\gt 38.$$

Since we are ultimately going to need a much larger value of $$n$$ (see below), the triangle inequality will automatically be satisfied for all triples once the acute-angled condition is met.

Now impose the acute-angled condition on the same extreme triple:

$$n^2+n^2 \gt (n+38)^2$$ $$\Longrightarrow 2n^2 \gt n^2+76n+1444$$ $$\Longrightarrow n^2-76n-1444\gt0.$$

Solve the quadratic inequality $$n^2-76n-1444=0$$:

Discriminant $$\Delta=76^2+4\cdot1444=5776+5776=11552,$$ $$\sqrt{\Delta}\approx107.46.$$

Roots $$n=\frac{76\pm\sqrt{11552}}{2}\approx\frac{76\pm107.46}{2}.$$ The positive root is $$\approx91.73$$.

Because $$n$$ must be an integer, the smallest feasible value is

$$n=92.$$

Verification:

For $$n=92$$ the extreme triple is $$92,92,130$$.

Triangle inequality: $$92+92=184\gt130.$$ Acute condition: $$92^2+92^2=2\cdot8464=16928\gt130^2=16900.$$

If the extreme triple is acute, every other triple from $$X$$ has either larger $$a$$ or $$b$$, or a smaller $$c$$, causing $$a^2+b^2-c^2$$ to increase. Hence all triples are acute when $$n\ge92$$, and they fail to be acute when $$n\le91$$ (because the triple $$n,n,n+38$$ becomes right- or obtuse-angled).

Therefore the smallest integer $$n$$ for which every choice of three elements from $$X$$ forms an acute-angled triangle is

92.