For how many numbers $$𝑛$$ in the set $$\left\{1,2,3,....,37\right\}$$ can we split the $$2𝑛$$ numbers $$1, 2, … , 2𝑛$$ into $$𝑛$$ pairs $$\left\{a_{i},b_{i}\right\},1 \le i \le n$$, such that $$\prod_{i=1}^{n}(a_{i}+b_{i})$$ is a square?

Let $$n$$ range over $$\{1,2,3,\dots ,37\}$$. For each admissible $$n$$ we must divide the integers $$1,2,\dots ,2n$$ into $$n$$ unordered pairs $$\{a_i,b_i\}$$ in such a way that the product $$\displaystyle\prod_{i=1}^{n}(a_i+b_i)$$ is a perfect square.

We treat three separate cases.

The only pairing is $$\{1,2\}$$ whose sum is $$3$$, so the product is $$3$$, which is not a square. Hence $$n=1$$ does not work.

Pair each integer with its “mirror image’’ about $$2n+1$$: $$\{1,2n\},\ \{2,2n-1\},\dots ,\{n,\,n+1\}.$$ Every pair has the same sum $$1+2n=2n+1,\;2+2n-1=2n+1,\;\dots ,n+(n+1)=2n+1.$$ Thus

$$\prod_{i=1}^{n}(a_i+b_i)=(2n+1)^n.$$

Because the exponent $$n=2m$$ is even, each prime factor of $$2n+1$$ occurs an even number of times, so the product is a perfect square. Therefore every even $$n$$ (that is, $$n=2,4,6,\dots ,36$$) satisfies the requirement.

Step I - fix the first six numbers.
Pair the numbers $$1$$ to $$6$$ as $$\{1,5\},\quad \{2,4\},\quad \{3,6\}.$$ The corresponding sums are $$6,\;6,\;9,$$ and their product is $$6\cdot6\cdot9=36\cdot9=18^2,$$ already a perfect square.

Step II - handle the remaining numbers.
After removing $$1$$ through $$6$$, the numbers left are $$7,8,9,\dots ,2n.$$ Their count is $$2n-6=2(2k+1)-6=4(k-1),$$ which is a multiple of $$4$$ because $$k-1$$ is an integer. Split them into consecutive blocks of four: $$\bigl(7,8,9,10\bigr),\ \bigl(11,12,13,14\bigr),\ \dots$$ In each block $$\bigl(x,x+1,x+2,x+3\bigr)$$ form the two pairs

$$\{x,x+3\},\qquad \{x+1,x+2\}.$$

Both pairs have the same sum $$x+(x+3)=x+1+(x+2)=2x+3.$$ Hence the contribution of this block to the overall product is $$(2x+3)\times(2x+3)=(2x+3)^2,$$ a perfect square.

Step III - combine everything.
The product from Step I is a square, and every four-number block from Step II contributes another square. The product of several perfect squares is again a perfect square, so the required condition is fulfilled for every odd $$n\ge 3$$ (that is, $$n=3,5,7,\dots ,37$$).

Combining all three cases:

• $$n=1$$ fails. • All other $$n\in\{1,2,\dots ,37\}$$ succeed.

Therefore the number of values of $$n$$ that work is $$37-1 = 36.$$

Final Answer: 36