Maximum of e

To maximize $e$ , lets minimize $a, b, c, d.$

Clearly, none of them can be $1$ .

Lets Set $a = 2$ & get smallest possible integers: $b, c, d$ in increasing order satisfying given equation.

$\frac{1}{2} + \frac{1}{b} + \frac{1}{c} +\frac{1}{d} +\frac{1}{e} = 1 \implies \frac{1}{b} + (\frac{1}{c} +\frac{1}{d} +\frac{1}{e}) = \frac{1}{2}$

Let $(\frac{1}{c} +\frac{1}{d} +\frac{1}{e} ) = \frac{1}{x} \implies \frac{1}{b} + \frac{1}{x} = \frac{1}{2}$

Rearranging, $2(b + x) = bx \implies (b - 2)(x - 2) = 4 = 1 \times 4.$

Set $(b - 2) = 1, (x - 2) = 4$ to minimize b. $\implies b = 3, x = 6 \implies (\frac{1}{c} +\frac{1}{d} +\frac{1}{e} ) = \frac{1}{6}$

Repeating same procedure, let $(\frac{1}{d} +\frac{1}{e} ) = \frac{1}{y}$ $( \frac{1}{c} +\frac{1}{y} ) = \frac{1}{6} \implies (c - 6)(y-6) = 36 = 1 \times 36$

Set $c = 7, y = 42 \implies \frac{1}{d} +\frac{1}{e} = \frac{1}{42} \implies (d-42)(e-42) = 1\times 1764$

Set $d = 43, e = 1806 \implies$ answer $\boxed{806}$