My Latest Answer
I have discovered a less trivial approximation for the growth rate of $f(n)$. But first I want to be rigorous about exactly what $f(n)$ is counting. I'm doing this in case I misinterpreted the question. My understanding is that $f(n)$ is the maximum number of distinct pairs $(i, j)$ with $1 \leq i < j \leq n$ such that there exists a set of $n$ positive integers $\{a_1, a_2, \dots, a_n\}$ with $a_i + a_j = s^2$ for some $s$. Clearly, $f(n) \leq {n \choose 2}$, since we can pick at most ${n \choose 2}$ pairs from $n$ objects.
Now, to find a lower bound for $f(n)$, we just need to find a set of pairs and a corresponding set of values, such that all values indexed by the pairs sum to perfect squares. Let's consider the set $\{1, 2, 3, \dots, n\}$ and define $h(n)$ as the number of distinct pairs of elements in this set that add to perfect squares. Clearly $h(n) \leq f(n)$. I will now put forth a very simple proof demonstration that:
$$n^{\frac32} \sim h(n)$$
Consider adding $n+1$ to the set $\{1, 2, 3, \dots, n\}$. We now form all the pairs $(1, n+1), (2, n+1), \dots, (n, n+1)$ whose corresponding sums are $n+2, n+3, \dots, 2n+1$. Now, let $s(a, b)$ denote the number of perfect squares in the interval $[a, b]$. Then we have:
$$h(n+1) = h(n) + s(n+2, 2n+1)$$
Now, how many perfect squares are in the interval $[n+2, 2n+1]$? Well, roughly there are:
$$\sqrt{2n+1} - \sqrt{n + 2} \approx \sqrt{2n} - \sqrt{n} = \sqrt{n}(\sqrt{2} - 1)$$
So, roughly speaking, we have:
$$h(n+1) \approx h(n) + \sqrt{n}(\sqrt{2} - 1)$$
Expanding out the recursion and factoring out the $(\sqrt{2} - 1)$ then we get:
$$h(n) \sim \sqrt{1} + \sqrt{2} + \dots + \sqrt{n} \sim \int_1^n \sqrt{x}\,dx \sim n^{\frac32}$$
My maths here is not up to my usual standard of rigour, but I think it's sound. This would imply that the growth rate of $f(n)$ is somewhere between $n^{\frac32}$ and $n^2$.
Update: We can arrive at a lower bound for $h(n)$ in a different way, avoiding the use of integration. Consider the set $\{1, 2, 3, \dots, n\}$. There are $\lfloor\sqrt{n}\rfloor$ perfect squares in this set. For each perfect square $k^2$, we can form it with $\lfloor\dfrac{k^2 - 1}{2}\rfloor \geq \dfrac{k^2}{2} - 1$ unique pairs:
$$(1, k^2 - 1), (2, k^2 - 2), \dots, (\lfloor\dfrac{k^2 - 1}{2}\rfloor, \lceil\dfrac{k^2 + 1}{2}\rceil)$$
These aren't all the squares that can be formed, but they are all possible, so they are valid for a lower-bound argument. Let $R=\lfloor\sqrt{n}\rfloor$ for ease of notation. A lower bound is then:
$$\sum_{i=1}^{R}\left(\dfrac{i^2}{2} - 1\right) = \frac12\sum_{i=1}^{R}i^2 - R = \dfrac{R(R+1)(2R+1)}{12} - R$$
Clearly this is of the order $n^{\frac32}$.
My Boring Original Answer
I will give a very trivial answer to $(2)$. We have the bound:
$$n - 1 \leq f(n)$$
To see why, choose $a_1 = 1$ and for all $i \neq 1$ choose $a_i = i^2 - 1$. Then we have that for all $i \neq 1$, $a_i + a_1 = i^2$. Since there are $n - 1$ values of $i \neq 1 \leq n$, we obtain the lower bound.
Update: Paying tribute to the commenter Crostul and the OP ComplexPhi, who have done more work than I, we can improve this lower bound to:
$$f(n) \geq f(k) + n - k$$
for any $k \leq n$.
Noting the discovery of the commenter Zander above, we have that $f(4) = 6$, and so we have the following bound:
$$f(n) \geq n + 2 \;\;\; \mathrm{where} \;\;\; (n \geq 4)$$
If your $n$ is odd, then the middle number has to be $50/n$. The odd divisors of $50$ are $1$ and $5$, which gives us two solutions $50=50$ and $8+9+10+11+12=50$
If $n$ is even, then $50/n$ is the half-integer between the middle two numbers. So $n$ has to be an even divisor of $100$, but not a divisor of $50$, so $n=4$ or $20$.
If $n=4$, then $50/4 = 12.5$ and we get $11+12+13+14=50.$
If $n=20$. then $50/20 = 2.5$ and we get $-7+-6+-5+\cdots +11+12 = 50.$
So there are 4 answers: $n=1, 4, 5, $ and $20$.
Edit: As Bill points out, I missed the divisors $25$ and $100$, which give two more answers: $50 = -10+-11+\cdots+14$ and $50 = -49 +-48+\cdots +50$.
Note that each solution with negative integers is related to an all-positive solution. From the solution $11+12+13+14=50$, we just prepend the terms $-10, -9, \ldots, 10$, which add to $0$, and we have another solution.
Best Answer
You can solve this simply by looking at the smallest number you can form and the largest number you can form. Let's say $\{a_1, a_2, ..., a_n\}$ is ascending. Then let $x = a_1 + a_2$ be the smallest number you can form. Let $y = a_{n-1} + a_n$ be the largest number you can form. Then you can see that you can form every number in between $x$ and $y$ (Hint: You can shift the sequence $\{a_1, a_2, ..., a_n\}$ so that it looks like $\{1, 2, ..., n\}$). Therefore, you can create $y - x + 1$ distinct numbers. Looking at our shifted values, we can create $(2n-1)-(3)+1=2n-3$ distinct numbers.