14. In a dormitory of a school, there are several students, one of whom serves as the dorm leader. During New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator. Each dormitory administrator also gives a greeting card back to the dorm leader. In this way, a total of 51 greeting cards were used. How many students live in this dormitory?
14. Let there be $x$ students in this dormitory, and $y$ administrators in the dormitory building $\left(x, y \in \mathbf{N}_{+}\right)$. According to the problem, we have
$$
x(x-1)+x y+y=51 \text{. }
$$

Simplifying, we get $x^{2}+(y-1) x+y-51=0$.
Thus, $\Delta=(y-1)^{2}-4(y-51)$
$$
=y^{2}-6 y+205=(y-3)^{2}+196 \text{. }
$$

Since $x \in \mathbf{N}_{+}$, $\Delta$ must be a perfect square.
Let $(y-3)^{2}+196=k^{2}(k \in \mathbf{N})$. Then
$$
(y-3+k)(y-3-k)=-196 \text{, }
$$

where $y-3+k$ and $y-3-k$ have the same parity, and $y-3+k \geqslant y-3-k$.
So, $\left\{\begin{array}{l}y-3+k=2, \\ y-3-k=-98\end{array}\right.$

or $\left\{\begin{array}{l}y-3+k=98, \\ y-3-k=-2\end{array}\right.$
or $\left\{\begin{array}{l}y-3+k=14, \\ y-3-k=-14 .\end{array}\right.$
From equation set (1), we get $y=-45$, which is not valid, so we discard it;
From equation set (2), we get $y=51$, in which case the original equation becomes $x^{2}+50 x=0$, solving for $x_{1}=-50, x_{2}=0$, both of which are not valid, so we discard them;

From equation set (3), we get $y=3$, in which case the original equation becomes $x^{2}+2 x-48=0$, solving for $x_{1}=-8$ (not valid, discard), $x_{2}=6$.
Answer: There are 6 students in this dormitory.