EDIT: See the updated question here. The technical assumptions in the question below were not adequate (as discussed in my answer below).
We follow the standard setting for Wald's equation.
Let $x_1, x_2, x_3, \ldots$ be a sequence of random variables.
Let random variable $N$ be a stopping time with $E[N] = \infty$. (Here we deviate from the standard assumptions!)
Suppose that, for all $n$, we have $E[x_n \,|\, N \ge n] = 1$, and, say, $x_n \ge -1000$.
My question: Do these assumptions imply that that $E[\sum_{n=1}^N x_n]$ is well-defined and infinite? If so, give a proof. If not, give a counterexample.
For intuition, consider the case that $x_n = 1$ for all $n$. Then the expectation of the sum in question equals $E[N]$, so is infinite by Assumption 2. The question is whether this continues to hold if, more generally, each $x_n$ is allowed to be a random variable with mean 1 (and uniformly bounded from below).
To show what would have to happen in any counterexample, consider the following sequence of identities, which can be used to show that $E[\sum_{n=1}^N x_n] = E[N]$ in the case that $E[N]$ is finite.
Replace the assumption $E[N]=\infty$ above with the assumption that $E[N]$ is finite. Then
$\begin{align} E\Big[\sum_{n=1}^N x_n\Big] &{} =~ \sum_{i=1}^\infty \Pr[ N=i ]\, E\Big[\sum_{n=1}^N x_n ~|~ N=i\Big] \\ &{} =~ \sum_{i=1}^\infty \sum_{n=1}^i \Pr[N=i] \, E[x_n \,|\, N=i] && \textit{(linearity of expectation)}~~~~~~~~~~(1)\\ &{} \mathbin{\stackrel{*}{=}}~ \sum_{n=1}^\infty \sum_{i=n}^\infty \Pr[N=i] \, E[x_n\, |\, N=i] && \textit{(holds given that $E[N]$ is finite; see below)} \\ &{} =~ \sum_{n=1}^\infty \Pr[N\ge i]\,E[x_n\,|\, N \ge i] \\ &{} =~ \sum_{n=1}^\infty \Pr[N\ge i] &&\textit{(as $E[x_n\,|\, N \ge i] = 1$)} \\ &{} =~ E[N]. \end{align}$
All equalities above except the third (marked $\stackrel{*}{=}$) follow from basic properties of probability. Given that $E[N]$ is finite, the third must also hold, because the total of the negative terms in the sum (1) is at least $$\sum_{i=1}^\infty \sum_{n=1}^i -1000 \Pr[N=i] = -1000 E[N],$$ so the total of the negative terms is finite, so changing the order of summation in (1) does not change the value.
The point is that, in any suitable counterexample for my question, the sum (1) in question must be finite, but changing the order of summation (as above) must make the value infinite.
Note also that (by a standard application of Wald's equation) for every $m$ we have that $E[\sum_{n=1}^{\min\{m, N\}} x_n] = E[\min\{m, N\}]$, which must tend to $\infty$ with $m$ (given that $E[N]=\infty$).
Answering my own question to point out that the technical assumptions given are not sufficient --- with just the assumptions as given, it can be that $E[\sum_{i=1}^N x_n]$ is not well-defined.
Consider the random experiment that sets $N=\infty$ with probability 1, then flips a fair coin and sets each $x_n = 3$ if the coin comes up heads, and otherwise sets each $x_n = -1$. Then $E[x_n | N \ge n] = 1$, but $\sum_{i=1}^N x_n$ is $\infty$ with probability 1/2, and $-\infty$ with probability 1/2, so the expectation is not well defined.
This is not a satisfying counterexample, though, because we implicitly had in mind the additional assumption that, for all $n$, $$E[x_n\, |\, x_1, x_2, \ldots, x_{n-1}] = 1.$$ That is, the expectation of $x_n$, conditioned on the event that $N\ge n$ and all the "previously observed" values of $x_1, x_2, \ldots, x_n$, should be 1. Such conditioning is usually assumed for Wald's equation.
I've created another question that incorporates this assumption.