How can I get Express.js to 404 only on missing routes?

You just need to put it at the end of all route.

Take a look at the second example of Passing Route Control:

var express = require('express')
  , app = express.createServer();

var users = [{ name: 'tj' }];

app.all('/user/:id/:op?', function(req, res, next){
  req.user = users[req.params.id];
  if (req.user) {
    next();
  } else {
    next(new Error('cannot find user ' + req.params.id));
  }
});

app.get('/user/:id', function(req, res){
  res.send('viewing ' + req.user.name);
});

app.get('/user/:id/edit', function(req, res){
  res.send('editing ' + req.user.name);
});

app.put('/user/:id', function(req, res){
  res.send('updating ' + req.user.name);
});

app.get('*', function(req, res){
  res.send('what???', 404);
});

app.listen(3000); 

Alternatively you can do nothing because all route which does not match will produce a 404. Then you can use this code to display the right template:

app.error(function(err, req, res, next){
    if (err instanceof NotFound) {
        res.render('404.jade');
    } else {
        next(err);
    }
});

It’s documented in Error Handling.

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