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Source. Theorem I, printed p. 382 (physical PDF p. 4 of the JSTOR scan, whose first page is a cover sheet); definitions (1)--(3) on pp. 380--382; proof in Sections 3--5, pp. 382--386, through Theorems II and III (p. 384); author's note on Takenouchi, p. 387. Read on the page images; the scan's text layer garbles the formulas.

Statement

For positive integers $x_1,\ldots,x_r$ define $f_r(x)$ by

$$\frac1{f_r(x)}=1-\frac1{x_1}-\frac1{x_2}-\cdots-\frac1{x_r}\qquad(3),$$

and let $u_1=1$, $u_{k+1}=u_k(u_k+1)$ (2), so the $u_k$ are $1,2,6,42,1806,\ldots$, each one less than the Sylvester numbers $2,3,7,43,1807,\ldots$

Theorem I (p. 382) reads: "The maximum finite value of $f_{n-1}(x)$ for all positive integral values of $x_1,x_2,\cdots,x_{n-1}$ is $u_n$ as defined by (2). There is but one set of $x$'s which gives this maximum value, namely that in which $x_k=u_k+1$, $k=1,2,\cdots,n-1$."

Equivalently: the least positive value of $1-\sum_{k=1}^{n-1}1/x_k$ over positive integers is $1/u_n$, attained only at $x_k=u_k+1$. Consequences: Kellogg's assertion that in any solution of $1=\sum_{i=1}^n1/x_i$ in positive integers the largest unknown is at most $u_n$ (since $x_n=f_{n-1}(x)$ by (1) and (3)); and, in the language of problem 206 (a deduction recorded here, not a statement of the paper), since the $x_k$ in Theorem I need not be distinct while the unique optimal set $x_k=u_k+1$ has distinct members, the best sum of $n-1$ distinct unit fractions below $1$ is $\sum_{k\le n-1}1/(u_k+1)=1-1/u_n$, so the best underapproximations of $1$ are nested and greedy for every $n$, not only eventually. Curtiss notes (p. 382) that, unlike Kellogg, he does not restrict to $x$'s making $f_{n-1}(x)$ an integer; the maximum turns out to be one.

Proof structure (pp. 382--386)

• Section 3 (p. 382): if all but the last $x$ are fixed, $f_{n-1}$ is largest finite when $x_{n-1}$ is the least integer exceeding $f_{n-2}(x)$, so a maximum needs $x_{n-1}=E(f_{n-2}(x))+1$ (4), $E$ the integer part; a set with this property for its largest member is compact, and a compact set with a single largest member is reduced. Two reduction procedures (pp. 383--384) turn any set with $1/f_r(x)>0$ into a reduced set, the first strictly increasing $f_r$; Theorem II (p. 384): a maximizing set must be reduced, so that, suitably labelled, $x_1\le\cdots\le x_{n-2}<x_{n-1}$ with (4); the proof assumes $n>2$, the case $n=2$ being noted as easy.
• Section 4 (pp. 384--386): for reduced sets $f_{n-1}(x)\le x_1\cdots x_{n-1}$ (10), hence $f_{n-1}(x)\le\phi_{n-2}(x)=x_1\cdots x_{n-2}[f_{n-2}(x)+1]$ (12) under $x_1\le\cdots\le x_{n-2}\le f_{n-2}(x)$ (11); Theorem III (p. 384): a set maximizing $\phi_{n-2}$ subject to (11) is reduced, proved by showing that each reduction step increases $\phi_{n-2}$.
• Section 5 (p. 386): iterating, $f_{n-1}(x)\le\phi_{n-2}(x)\le\phi_{n-3}(x)[\phi_{n-3}(x)+1]\le\cdots\le U_{n-2}$ with $U_1=\phi_1(x)$ and $U_{k+1}=U_k(U_k+1)$; the constraint $x_1\le f_1(x)=x_1/(x_1-1)$ forces $x_1=2$, so $U_1=6=u_3$ and $U_{n-2}=u_n$; the value $u_n$ is attained at $x_k=u_k+1$, and the necessary conditions at each reduction leave only that set.
• Author's note (p. 387): Takenouchi, in a paper that had then just appeared (Proc. Phys.-Math. Soc. Japan (3) 3, 78--92), treated $\sum1/x_i=b/a$ and, for $a=(m+1)b-1$, found the maximum unknown $A_n$ ($n>1$) with $A_1=m$, $A_2=a(A_1+1)$, $A_{k+1}=A_k(A_k+1)$, which for $b=m=1$ is Kellogg's theorem; Curtiss states that Theorem I itself is not proved there and states, for $b\le a$ and $n>1$, an upper bound $B_n$ for the maximum finite value of the analogous $f_{n-1}(x)$, reached when $a=(m+1)b-1$.

The steps were read for structure on the page images and are recorded as a sketch; the proof is not rewritten in full and has not been independently reviewed.

Read depth

Claims checked (Theorem I and definitions (1)--(4) read clause by clause on the page images of pp. 380--382); proof read for structure.

Bears on. #206: the case $x=1$, where by the deduction above (not a statement of the paper), the best sums of $n$ distinct unit fractions below $1$ are greedy at every length; it says nothing about other $x$.