The problem of finding the smallest positive $a,b,c$ for which $a/(b+c)+b/(a+c)+c/(a+b)=4$ turns out to be surprisingly difficult, and has made the rounds on the internet and social media, and Andrew Bremner and Allan Macleod have written a paper on it. Looking at the associated elliptic curve, you can deduce the smallest solution is \begin{align} a &= 154476802108746166441951315019919837485664325669565431700026634898253202035277999\\ b &= 4373612677928697257861252602371390152816537558161613618621437993378423467772036 \\ c &= 36875131794129999827197811565225474825492979968971970996283137471637224634055579 \end{align} and permutations. Another interesting question is to find the smallest $a,b,c$ (in some sense) for which $a/(b+c)+b/(a+c)+c/(a+b)$ is an integer $>2$ ($a=1,b=1,c=3$ satisfy $N=2$). I believe it may be \begin{align} a &= 6383088000457968550863626020707964592827 \\ b &= 86901761472912010754925122912564100123 \\ c &= 2743260400516683056616306684496286550899825 \end{align}

which yield the value $N=424$. One of the reasons these numbers are so small is that the associated elliptic curve has rank 3, hence you have a great deal of flexibility in combining the generators in order to get a point in the region of the non-identity component that maps to positive $a,b,c$. I'm not certain about the smallest value of $N$ associated with a rank 4 elliptic curve, but it may be $N=13502$.

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    $\begingroup$ I computed with Magma the 2-Selmer groups of these curves (maybe under GRH) and the smallest $N$ such that $E$ has rank 4 indeed seems to be 13502 (I restricted to those curves with root number $+1$ so the parity conjecture is also used). There is a conjecture of Lang asserting that the size of the smallest non-torsion point of an elliptic curve is bounded from below by an absolute constant times the size of the minimal discriminant (which here is polynomial in $N$). $\endgroup$ – François Brunault Nov 22 '18 at 22:41
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    $\begingroup$ I finished my rank 4 code, and it found that N=13502 has lowest height solution with max(a,b,c) having 1322 digits, which is quite a bit larger than the above solution. The code and calculation are here: github.com/octonion/puzzles/blob/master/twitter/fruits/… $\endgroup$ – Christopher D. Long Nov 23 '18 at 3:04

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