We first prove the following lemma.

Lemma. Let $f(x)$ be a polynomial with complex coefficients such that its leading coefficient is rational. If for some positive integer $k$ we have $f(x)^k \in \mathbb{Z}[X]$, then the polynomial $f(x)$ would also be in $\mathbb{Z}[X]$.

Proof. Assume $f(x) = a_n x^n + \cdots + a_0$. Then we can write $f(x)^k = b_k x^{kn} + \cdots + b_1 x + b_0$. Now, assume inductively that $a_n, a_{n-1}, \ldots, a_{n-s}$ are rational then we must prove $a_{n-s-1}$ is rational. For proving this, we compare the coefficient of $x^{kn-s-1}$ in the both sides. Then we can say that $b_{kn-s-1} = a_{n-s-1}a_n^{k-1} + S$ where $S$ is a rational number. We conclude that $a_{n-s-1}$ is also a rational number. That is, $f(x)$ must have rational coefficients.

Now we want to prove that $f(x)$ has integer coefficients. If for some $k$ we have $f(x)^k = f(x)^{k-1} \cdot f(x) \in \mathbb{Z}[X]$, then by use of Gauss's lemma, we can say that there exists a rational number $q$ such that the polynomials $qf(x)$ and $q^{-1}f(x)^{k-1}$ have integer coefficients. Assume $q = \frac{a}{b}$ where $\gcd(a, b) = 1$. Thus we can say that $af(x)$ and $bf(x)^{k-1}$ have integer coefficients. Therefore, the polynomial $a^{k-1}f(x)^{k-1}$ has integer coefficients. We know there exist integers $s, t$ such that $a^{k-1}s + bt = 1$. Thus the polynomial $a^{k-1}sf(x)^{k-1} + btf(x)^{k-1} = f(x)^{k-1}$ has integer coefficients. Continuing this process we can observe that $f(x)$ has integer coefficients.

We also need the following lemma.

Lemma. Let $P(x)$ and $Q(x)$ be two polynomials with rational coefficients such that $Q(x)$ divides $P(x)$ then the quotient of these polynomials, i.e., $R(x) = \frac{P(x)}{Q(x)}$ has rational coefficients.

Proof. It is clear based on the division algorithm of polynomials.

Back to our problem, let $R(x)$, $S(x)$ and $T(x)$ be three elements of the desired set. Since $R(x)S(x)$, $R(x)T(x)$ and $S(x)T(x)$ all have integer coefficients then $R(x)^2 = \frac{R(x)S(x)R(x)T(x)}{S(x)T(x)}$ has also rational coefficients. Let $r, s$ and $t$ be the leading coefficients of $R(x)$, $S(x)$ and $T(x)$, respectively. Since $R(x)S(x)$, $S(x)T(x)$ and $R(x)T(x)$ are all special, it follows that $rs = st = tr = 1$. Hence, $r = s = t = \pm 1$. Applying above lemma, we find that $R(x)$, $S(x)$ and $T(x)$ have rational coefficients. Thus, $R(1)$, $S(1)$ and $T(1)$ are rational numbers. Further, $R(1)T(1) = R(1)S(1) = T(1)S(1) = 1399$. It follows that $R(1), S(1), T(1) \in \{\pm\sqrt{1399}\}$. This is impossible. So, such a set doesn't exist. ■