Solution:

Write $a = BC = 219,$ $b = CA,$ $c = AB,$ and $s$ for the semiperimeter. The splitting line through $M$ meets $\overline{BC}$ at the point $X$ with $BX = s - \frac{c}{2}$ (then each piece has perimeter $s$). In triangle $BMX,$ the law of sines shows $\angle BXM = \frac{C}{2}:$ this needs $c \sin\left(B + \frac{C}{2}\right) = (a + b)\sin\frac{C}{2},$ which reduces via $a + b = 2R(\sin A + \sin B) = 4R\cos\frac{C}{2}\cos\frac{A - B}{2}$ and $c = 4R \sin\frac{C}{2}\cos\frac{C}{2}$ to $\sin\left(B + \frac{C}{2}\right) = \cos\frac{A - B}{2},$ true because those angles are complementary. Hence the splitting line through $M$ is parallel to the angle bisector from $C,$ and likewise the one through $N$ is parallel to the bisector from $B.$

The internal bisectors from $B$ and $C$ meet at $90^\circ + \frac{A}{2} \gt 90^\circ,$ so the acute angle between the two splitting lines is $90^\circ - \frac{A}{2} = 30^\circ,$ forcing $\angle A = 120^\circ.$ The law of cosines gives $$219^2 = b^2 + c^2 + bc = (b + c)^2 - bc.$$ Set $p = b + c,$ so $bc = p^2 - 219^2$ and $b, c$ are roots of $t^2 - pt + (p^2 - 219^2),$ requiring $4 \cdot 219^2 - 3p^2$ to be a perfect square $k^2.$ Then $3 \mid k$ and $3 \mid p;$ writing $p = 3r$ and $k = 3m$ turns the condition into $m^2 + 3r^2 = 146^2.$ The triangle inequality $p \gt 219$ and $4 \cdot 219^2 \ge 3p^2$ restrict $74 \le r \le 84,$ and checking these, only $r = 80$ works, with $m = 46.$

So $b + c = 240$ and $bc = 240^2 - 47961 = 9639,$ giving $\{b, c\} = \{51, 189\}$ — a valid triangle. The perimeter is $219 + 240 = 459.$