1. Let the interior angles of the $n$-sided polygon be in A.P.:

$a,\ a+6,\ a+12,\ \dots,\ a+(n-1)6$

Since the largest interior angle is $219^\circ$, we have

$a+(n-1)6=219$

So,

$a=219-6(n-1)=225-6n$

2. The sum of interior angles of an $n$-gon is

$(n-2)180^\circ$

Also, sum of the A.P. is

$\frac{n}{2}(\text{first angle} + \text{last angle})$

Hence,

$\frac{n}{2}(a+219)=(n-2)180$

Substitute $a=225-6n$:

$\frac{n}{2}(225-6n+219)=(n-2)180$

$\frac{n}{2}(444-6n)=180n-360$

$n(222-3n)=180n-360$

$222n-3n^2=180n-360$

$42n-3n^2+360=0$

$3n^2-42n-360=0$

Divide by $3$:

$n^2-14n-120=0$

Factorizing,

$n^2-24n+10n-120=0$

$n(n-24)+10(n-24)=0$

$(n-24)(n+10)=0$

Since $n>0$,

$n=24$

3. Verification:

First angle:

$a=225-6(24)=225-144=81^\circ$

Last angle:

$81+23\cdot 6=81+138=219^\circ$

Sum:

$\frac{24}{2}(81+219)=12\cdot 300=3600^\circ$

For a 24-gon,

$(24-2)180=22\cdot 180=3960^\circ$

This seems inconsistent, so let us recheck the arithmetic.

From

$\frac{n}{2}(444-6n)=(n-2)180$

we simplify the left side correctly:

$\frac{n}{2}(444-6n)=n(222-3n)$

Thus,

$222n-3n^2=180n-360$

$42n-3n^2=-360$

$3n^2-42n-360=0$

Divide by $3$:

$n^2-14n-120=0$

$(n-20)(n+6)=0$

Therefore,

$n=20$

4. Final check:

First angle:

$a=219-6(19)=219-114=105^\circ$

Angles are:

$105^\circ, 111^\circ, 117^\circ, \dots, 219^\circ$

Sum of these 20 angles:

$\frac{20}{2}(105+219)=10\cdot 324=3240^\circ$

Sum of interior angles of a 20-gon:

$(20-2)180=18\cdot 180=3240^\circ$

This matches perfectly.

Hence, the polygon has $20$ sides.