If the quadrilateral has sides $a,b,c,d$ and perimeter $s = a+b+c+d$ then $a < b+c+d$. Also, $a$ divides $b+c+d$, so $b+c+d$ is at least twice $a$. But $b+c+d = s-a$, so $a$ also divides $s$, and $s$ must be at least $3$ times $a$, say $s=ka$ with $k\geqslant3$. Similarly, $s = lb$, $s=mc$ and $s=nd$, with all of $l,m,n$ greater than or equal to $3$.
Then $a = \dfrac1ks$, $b = \dfrac1ls$, $c = \dfrac1ms$, $d = \dfrac1ns$. Therefore $s = a+b+c+d = \dfrac1ks + \dfrac1ls + \dfrac1ms + \dfrac1ns$ and so $$\frac1k + \frac1l + \frac1m + \frac1n = 1.$$ But under the given conditions, if $a,b,c,d$ are all different then so are $k,l,m,n$, and the largest possible value for $\dfrac1k + \dfrac1l + \dfrac1m + \dfrac1n$ is $\dfrac13 +\dfrac14 + \dfrac15 + \dfrac16 = \dfrac{57}{60}$, which is less than $1$. So those conditions cannot be satisfied and therefore at least two of the four sides must have the same length.