Another answer for a completely different approach that has more potential to produce a correct answer in high dimensions.
I created a high density lattice $L$ in dimension 97 from the identity with the following first row (vectors are columns):
[1849600401647997099719919617918, 77491606716638510245109537658, 1525976942917100825918475963, 74020030346105712528538189620, 51941909136916852113082350389, 14011368459601438276486807412, 50822939581149515856915735546, 2509725539352575690590079057, 71748135378136038959690456165, 44171810008920839742209859013, 32719850946638740383177195077, 30023966707232726541396407076, 4410448856566107235653959856, 22022346329990180543792971711, 7269505503990621246930936020, 5107569876022684367034274130, 47778510178448380848329851693, 3169027828311763162013444590, 57480237440919598172784764429, 13180659738586100049888825360, 60213343575174306448588865612, 65767716966639197940060401284, 32829858938231484496185322443, 62920352240097256280791858782, 68771691370698613819098072377, 1189446426775895534566399888, 15622321082446447641269029228, 44854197141206405708649394123, 18743069471287551089559748119, 78100141274951600010628631304, 41336653782179264170800348173, 28886061428722075757026880056, 10430320172190962448775321281, 23275291645042566655909933981, 9052772967664433789458201405, 75118661056395305373457649071, 36024369475873917291385515901, 24329686458049949630078669272, 21999841835292501255568512142, 35735166411960669882213981372, 37822428212473917877193850830, 21508264293370451672165513701, 11674532436606371783473819749, 6242499836401929922573545309, 44310331499846649220024243828, 52015012976023730286199334013, 9946920677622169606617681376, 47598058001625475755489112483, 57667202535436450251667859446, 67120509210736141602830685965, 62099852611535925832337479401, 20379832896127026776968623871, 56578972203046231525042405448, 52003353802918059453785164051, 56586003582175107572782034325, 77415032294150861160530555875, 27716827577183024283562170388, 56838383954420862660792615481, 77244740436710851110420556574, 34907304650256412809700886291, 64664781684296567841076011641, 47531215060867292720500337851, 24488984341820349659649621937, 48259263501801547468879409155, 68079961159219693040054958499, 70346145409960265348894545256, 74735588078890947855987725542, 47998837475838422523126923115, 58202334552136529552226186712, 356488283196519720283167114, 34379228951744427157065767394, 19288837674944009785234496912, 19815697624832173898316400583, 22888127129397336162600462702, 12402340382435329901966351103, 7343124368453544417346578155, 47945165517372728594283927672, 22348591021805731698927620947, 48204981766234374553968627534, 30159946581970181385152488904, 10900707776910254490250210209, 15260134540509522683945503275, 27948496473361408358754701178, 48182739123020986961217661572, 35980702830244507680285631127, 70230630306463989263728556892, 56047883827612733579048229880, 9623951329583261678855132961, 55680584783950871236564644585, 60661234139611717690479236612, 54823003493318948369949365698, 31413576619803196280952014808, 35338249047649796737419966138, 31287456730543657901971615872, 78834491773686153881906217163, 53375041532170131381123655147, 36830407022141621320829586473]
(each of these has about 96 bits, this gives density approx 1.0, the hardest case to e.g. solve exact 0/1 knapsack, which was my original motivation).
I then used the gpu-optimized version of G6K of Wessel Van Woerden (see here) repeatedly to find as good a basis for $L$ as possible. On 4 NVIDIA A100s I was able in about 2 hours of realtime to find a basis with 1,4,18,73,1 vectors resp. of $\ell_2$-norm 24,25,26,27,99 resp. This is not Minkowski-reduced, but is considerably better than e.g. LLL can do (or even BKZ in reasonable time). I seemed to have to pick one large vector to ensure the basis had full rank.
I had the idea to look for vectors closest to `reasonably small' vectors in the positive orthant.
To find the closest vectors I used the Kannan embedding technique, which converts an SVP solver to a CVP solver in one dimension higher (see e.g. here).
Using this I picked fairly randomly the following non-lattice vector
$$(2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 2, 2, 3, 3, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 3, 3, 3, 2, 2, 2, 3, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 3, 2)^\top$$
and again with Van Woerden's implementation I found the following actual lattice vector in the positive orthant:
$$(2, 3, 4, 2, 1, 2, 4, 3, 2, 3, 1, 2, 2, 3, 4, 1, 3, 1, 2, 2, 3, 2, 2, 2, 1, 2, 3, 2, 3, 1, 3, 1, 3, 3, 3, 1, 3, 2, 4, 6, 3, 1, 2, 3, 3, 2, 1, 2, 4, 4, 1, 3, 4, 2, 2, 4, 2, 2, 2, 3, 3, 3, 1, 3, 3, 4, 1, 2, 2, 4, 2, 1, 3, 2, 1, 4, 3, 1, 3, 4, 6, 4, 2, 3, 3, 3, 1, 3, 3, 1, 2, 3, 3, 3, 4, 3, 1)^\top$$
in almost no time at all (a few seconds). Since the individual CVP instances can be solved so quickly, it opens the possibility to find many more short vectors in the positive orthant by running e.g. 1000s of CVP instances for random small positive non-lattice vectors.
I'm not 100% sure the good basis was crucial, but technically in the Kannan embedding we need the last coordinate to be 1 and this doesn't always get returned by the 'pumping' mechanism of the gpu SVP solver, but you can access the top $n$ (dimension) lifts and at least one of these is guaranteed to have this 1 coordinate because the top $n$ lifts are a basis for $L$.
I would really welcome any comments or ideas on how to improve this approach to find the shortest vector in the positive orthant. This would solve 0/1 knapsack (in exponential time of course but it would make many more dimensions feasible with gpu acceleration).