Let's approach the problem one variable at a time. Without loss of generality, assume that $J < K < L < M < N$.
What is J?
If $J = 1$, then we would have $\frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{KLMN} = 0$, which is clearly impossible. So $J \ne 1$.
If $J ≥ 4$, then the greatest the LHS could possibly be is $\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{4⋅5⋅6⋅7⋅8} = \frac{1189}{1344} < 1$. And increasing any variable simply makes a smaller fraction. It will always be less than 1. So, any solution with $J ≥ 4$ is ruled out.
OTOH, $J = 3$ produces an upper bound of $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{3⋅4⋅5⋅6⋅7} = \frac{551}{504} > 1$, which is OK.
So, $J \in \lbrace 2, 3 \rbrace$.
What is K?
Since there are only two possibilities for $J$, let's plug in each of them.
- If $J = 2$, then $\frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{2KLMN} = \frac{1}{2}$. As before, the LHS is maximized by taking all the variables to be consecutive integers.
- If $K = 6$, then we have $\frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{2⋅6⋅7⋅8⋅9} = \frac{3301}{6048} > \frac{1}{2}$, which is fine.
- But if $K = 7$, we have $\frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{2⋅7⋅8⋅9⋅10} = \frac{4829}{10080} < \frac{1}{2}$, which is too low. So $K ≤ 6$.
- Recalling that $K > J$, this means $K \in \lbrace 3, 4, 5, 6 \rbrace$.
- If $J = 3$, then $\frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{3KLMN} = \frac{2}{3}$.
- If $K = 4$, then the upper bound on the LHS is $\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{3⋅4⋅5⋅6⋅7} = \frac{383}{504} > \frac{2}{3}$, which is OK.
- But if $K = 5$, then we have $\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{3⋅5⋅6⋅7⋅8} = \frac{457}{720} < \frac{2}{3}$, which is too low.
- So $K = 4$ is the only possibility.
Taking the union of the cases, we have $K \in \lbrace 3, 4, 5, 6 \rbrace$.
What is L?
From the previous section, we have 5 possibilities for $(J, K)$:
- $J = 2$, $K = 3$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{6LMN} = \frac{1}{6}$, and $4 ≤ L ≤ 17$.
- $J = 2$, $K = 4$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{8LMN} = \frac{1}{4}$, and $5 ≤ L ≤ 11$.
- $J = 2$, $K = 5$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{10LMN} = \frac{3}{10}$, and $6 ≤ L ≤ 9$.
- $J = 2$, $K = 6$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{12LMN} = \frac{1}{3}$, and $7 ≤ L ≤ 8$.
- $J = 3$, $K = 4$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{12LMN} = \frac{5}{12}$, and $5 ≤ L ≤ 6$.
Taking the union of these gives $4 ≤ L ≤ 17$.
What is M?
If we take the minimum values for the other variables: $J = 2$, $K = 3$, and $L = 4$, then $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{M} + \frac{1}{N} + \frac{1}{24MN} = 1$, or $\frac{1}{M} + \frac{1}{N} + \frac{1}{24MN} = \frac{-1}{12}$. That negative number on the right means that the approach used to find an upper bound for J, K, and L won't work for M. So, let's just skip it and come back to it later.
What is N?
If we have values for the other 4 variables, then we can solve for N directly.
$$\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{JKLMN} = 1$$
$$\frac{1}{N}(1 + \frac{1}{JKLM}) = 1 - (\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M})$$
$$\frac{1}{N} = \frac{1 - (\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M})}{1 + \frac{1}{JKLM}}$$
$$N = \frac{1 + \frac{1}{JKLM}}{1 - (\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M})}$$
$$N = \frac{JKLM + 1}{JKLM - (KLM + JLM + JKM + JKL)}$$
All we have to do is confirm that this number is an integer, and that it is greater than $M$.
Brute force
A slight modification of ab123's Python script to use my tighter bounds for J, K, and L; and formula for N.
from fractions import Fraction
MAX_M = 1000000
for J in range(2, 4):
for K in range(J + 1, 7):
for L in range(K + 1, 18):
for M in range(L + 1, MAX_M + 1):
N1 = J*K*L*M + 1
N2 = J*K*L*M - (K*L*M + J*L*M + J*K*M + J*K*L)
if N2 != 0:
N = Fraction(N1, N2)
if N.denominator == 1 and N > M:
print(J, K, L, M, N)
This gives three solutions:
- (2, 3, 7, 43, 1807)
- (2, 3, 7, 47, 395)
- (2, 3, 11, 23, 31)
Perhaps other solutions exist with $M > 10^6$. Or someone can prove that they don't.