= Solution
The coefficient matrix has <eigenvector> $u_+=(1,1)^T$ with eigenvalue $3$ and eigenvector $u_-=(1,-1)^T$ with eigenvalue $-1$. Write
$$
y(t)=p(t)u_++q(t)u_-.
$$
Since
$$
\begin{pmatrix}e^{2t}\\-2t\end{pmatrix}
=\frac{e^{2t}-2t}{2}u_+
+\frac{e^{2t}+2t}{2}u_-,
$$
and $y(0)=(1,-2)^T=-\tfrac12u_++\tfrac32u_-$, the system diagonalises to
$$
p'=3p+\frac{e^{2t}-2t}{2},\qquad p(0)=-\frac12,
$$
$$
q'=-q+\frac{e^{2t}+2t}{2},\qquad q(0)=\frac32.
$$
Using an <integrating factor> gives
$$
p(t)=-\frac19e^{3t}-\frac12e^{2t}+\frac t3+\frac19,
$$
$$
q(t)=\frac16e^{2t}+t-1+\frac73e^{-t}.
$$
The unique fastest term is therefore
$$
y(t)=-\frac19e^{3t}u_++O(e^{2t}).
$$
Consequently $e^{-nt}y(t)$ diverges for $n<3$, tends to zero for $n>3$, and for $n=3$ tends to the finite nonzero vector
$$
\boxed{\lim_{t\to\infty}e^{-3t}y(t)
=-\frac19\begin{pmatrix}1\\1\end{pmatrix}}.
$$
Thus the only requested integer is
$$
\boxed{n=3}.
$$
This is an instance of <dominant eigenmode in a forced linear system>.
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