Imagine the following array:
$$\begin{array}[cccccccccc] .1 && 2 && 3 && 4 && \ldots && n-2 && n-1 && n \\ 2n && 2n-1 && 2n-2 && 2n-3 && \ldots && n+3 && n+2 && n+1\end{array}$$
Notice that each column sums to $2n+1$ and all of the numbers from $1$ to $2n$ are used in the array. There are $n$ columns. What you want to prove is that if you were to highlight $n+1$ numbers in this array (i.e. the elements of $T$), there would be a whole column highlighted, and that pair would sum to $2n+1$.
The pidgeonhole principle essentially says that we cannot possibly highlight $n+1$ numbers such that no two lie in the same column, if there are but $n$ columns. If you want to see this, then just take a small array, like for $n=3$:
$$\begin{array}.1 && 2 && 3\\6 && 5 && 4\end{array}$$
Now, let's start highlighting some numbers, trying to avoid putting two in a column. Our goal is to highlight $4$ numbers, as that is the size of the set $T$. We could start by putting $1$ in $T$:
$$\begin{array}.\color{red}1 && 2 && 3\\6 && 5 && 4\end{array}$$
but now we know we can't put $6$ in $T$ too, because that would sum to $2n+1$. So we might choose $5$ as our next number, forbidding $2$ and we might choose $4$ as the number after that:
$$\begin{array}.\color{red}1 && 2 && 3\\6 && \color{red}5 && \color{red}4\end{array}$$
So, now we have a highlighted number in every column- and adding any further number to the set $T$ would create a pair summing to $7$. But this means that we can't have a fourth element in $T$, at least given how we started - and the pidgeonhole principle guarantees that we can never choose a set of size $4$ without putting two elements in one column.
The key point here is that we should imagine that, as we're creating $T$, we're not choosing numbers to put in it, we're choosing which column to take the numbers from. There are $n$ columns, and we need to make $n+1$ choices - thus we will, at some point, choose the same column twice, and in this context, that means we need to have both elements of some column in $T$, and this forms a pair summing to $2n+1$.
The answer is 16: to show that 15 is possible, just take $2,3,5,7,\ldots,47$.
Now for the fun part, making the pigeonholes. The first pigeonhole is $$S_1=\{2,4,6,8,\ldots\}$$
and the second is $$S_2=\{3,9,15,\ldots\}$$
Continuing thus, the $i$th set is (for $i>1$)
$$S_i=\{p_i,\ldots\}$$ where $p_i$ is the $i$th prime, and the elements that occur in the set are multiples of $p_i$ which don't occur in previous sets.
Note that every positive integer ends up in one of these sets; in particular, the numbers 2-50 end up in the first 15 of these sets. Thus taking $16$ numbers guarantees two in the same set (by PHP) and we're done, since any two numbers in a set trivially share a factor.
Best Answer
It could be interesting to show the more general case, that is, if we know the prime divisors of all elements in $M$ are amongst $p_{1}, p_{2}, ..., p_{n}$, and $M$ has at least $2^{n} \cdot 3 + 1$ elements, then it contains at least one subset of four distinct elements whose product is a fourth power.
The trick is to assign each element $m$ in $M$ with an $n$-tuple $(x_{1}, x_{2}, ..., x_{n})$, where $x_{i}$ is an indicator variable equal to $0$ if the exponent of $p_{i}$ in $m$'s prime factorization is even, and $1$ otherwise.
As my teacher explained this problem, these tuples become our objects with which to consider the pigeonhole principle, and our boxes are the possible choices of $0$'s and $1$'s for each indicator. Then by the pigeonhole principle, we have that every subset of our $2^{n}+1$ elements of $M$ contains two (distinct) elements with the same object, or $n$-tuple, and consequently the product of these two elements is a square.
Our process consists of repeatedly removing these pairs, and replacing them with two of our remaining possible numbers. Since $M$ has at least $2^{n}\cdot 3 + 1$ elements, we can select at least $2^{n}+1$ pairs.
Now we must consider the $2^{n}+1$ numbers that are products of the two elements of each pair. We can repeat the same argument as above once again, to have four elements $a,b,c,d$ in $M$ where $\sqrt{ab}\sqrt{cd}$ is a perfect square. Then $abcd$ is a fourth power, and our result has been shown.
In your case, we have that $n=9$, as $1985 > 3 \cdot 2^{9} + 1 = 1537$.