Finding a solution for this system of two diophantine equations (depending on a parameter) I propose the following problem (Maybe it has a trivial solution):
Let $n$ be a positive integer such that $$n\equiv1 \pmod 4.$$
Then the problem is to find a rational $x$ as a function of $n$ such that 
$$ \dfrac{3n+3x+n^{2}}{12} \quad\text{and}\quad \dfrac{n(n+3)(3n+3x+n^{2})}{36x}$$
are both strictly positive integers. Or at least how one can proves that such a $x$ exists.
 A: One can use common subexpressions to get a simple answer.  Note that $a=n(n+3)$ must be a multiple of $4$. Setting $y=3x$, we look for $a+y$ is a multiple of $12$ and $a(a+y)$ is a multiple of $12y$.  If we can pick $y$ to meet $a+y$ is a multiple of $12$, then $y$ is an integer and it suffices to also pick $y$ being a divisor of $a$.  This is possible for $n=5$ and in general $y$ being a $2 mod 3$ divisor of $n(n+3)$ should work.  I find $x=(n+3)/3$ works for many $n$.
Edit 2018.01.16
The problem is a little more intriguing.  If $n$ is $0$ or $2 \equiv1 \pmod 3$, then letting $3x$ be $n+3$ gives a solution as can be easily checked.  For $n=1 \equiv1 \pmod 3$ (and so $n$ is $1 \pmod {12}$, $3x$ needs to be an integer which is $\equiv8 \pmod {12}$ to satisfy the first relation, and setting $y=3x$ and $a=n(n+3)$ gives $12y$ has to divide $a(a+y)$. This is easy if $y$ divides $a$, while if $y$ does not divide a then with $d$ being the greatest divisor of $a$ and $y$, and $b =y/d$, we get $b$ has to divide $d$. For small $n$ with $n$ being $\equiv1 \pmod {12}$, we can find such a divisor $y$ which divides $a$, but it is not clear that we can always do that. So far $y=8$ or $20$ works for small $n$.
End Edit 2018.01.16
Gerhard "Addition More Complex Than Multiplication?" Paseman, 2018.01.16.
A: At first glance I thought that for $(1)$ to have solutions $n$ had to be a multiple of $3$, which is wrong  when $x$ can be a rational, as Gerhard pointed out. So what follows is a solution when $n=3m$ for some $m$. Remain to be treated the cases $n\equiv1 \pmod 3$ and $n\equiv2 \pmod 3$.
So $n=3m$, with $m\equiv3 \pmod 4$.
Writing $m=4p+3$ condition $(1)$ becomes
$$\dfrac{3(4p+3)+x+3(4p+3)^2}{4}\in \mathbb{N}$$
which is equivalent to $$\dfrac{1+x+3}{4}\in \mathbb{N}$$
so that $$(1)\Longleftrightarrow x\in 4\mathbb{N}$$.
Now if you plug in $n=3(4p+3)$ and $x=4q$ into $(2)$, you get massive simplifications into
$$\dfrac{(4p+3)(p+1)(3q+9(4p+3)(p+1))}{q}\in \mathbb{N}$$
showing that for instance any factor of $(4p+3)(p+1)$ will do.

Conclusion: when $n\equiv0 \pmod 3$
  
  
*
  
*$p=\dfrac{n-9}{12}$ is an integer, 
  
*$x=4(p+1)$ is a solution.
  

