This writeup uses the notation of Sierpinski's Theorem, and anyway is a proof of what's there, so why don't you read that first?
Suppose, to the contrary, that for some a_{1},...,a_{k},b there exist infinitely many solutions in natural numbers to the equation (*). Let F be the set of all solutions. By Robinson's overspill lemma, F^{^} (see NSA: some pseudo set theory for an explanation of "content") contains non-standard solutions of the equation. These are solutions where at least one `x`_{j} is non-standard: call N the set of natural numbers; the pseudo-elements N^{^}-N are the "non-standard natural numbers".

Let's examine one such non-standard solution. It claims that a_{1}/`x`_{1}+...+a_{k}/`x`_{k}=b, where some of the `x`'s are non-standard natural numbers.

#### Claim 1:

If `x` is a non-standard natural number, then `x`>n for any (standard) natural number n.
#### Proof of claim 1:

There are only finitely many (standard) natural numbers {`t`: `t`<=n}; by Robinson's overspill lemma, there are no such non-standard ones. Since ">" (on the real numbers, say) is a total order, a property we can define in first order predicate calculus, "*->" (the transferred predicate on the pseudo-real numbers) is a total order too.

It directly follows that for any (positive) real number a, the *-number a/`x` is positive and smaller than any (standard) real number ε>0. Call such a number a positive infinitesimal.

#### Claim 2:

The sum of `j` (>0) positive infinitesimals is a positive infinitesimal.
#### Proof of claim 2:

In the standard world, it's true that for all `x`_{1},...,`x`_{j} and ε/`j`, if 0<`x`_{1},...,`x`_{j}<ε/`j` then 0<`x`_{1}+...+`x`_{j}<ε. Transferring to the non-standard world, taking ε to be standard real numbers, shows the claim.

We've seen that a non-standard solution contains some non-standard elements. Suppose

without loss of generality that

`x`_{1},...,

`x`_{j} are non-standard and

`x`_{j+1},...,

`x`_{k} are standard (0<

`j`<=

`k`). Then

a_{1}/`x`_{1}+...+a_{j}/`x`_{j} = b - (a_{j+1}/`x`_{j+1}+...+a_{k}/`x`_{k}).

Now the

LHS is a positive infinitesimal, but the

RHS is some expression composed entirely of standard real numbers! Thus the RHS is some standard real number ε. It cannot equal a positive infinitesimal: it's positive, so it cannot be less than ε/2. But ε/2 is a standard real number too, thus larger than any positive infinitesimal.

This contradiction shows that the premise (having an infinite number of solutions) is absurd!