The simplest solution is indeed to take N random values and divide by the sum.
A more generic solution is to use the Dirichlet distribution
which is available in numpy.
By changing the parameters of the distribution you can change the "randomness" of individual numbers
>>> import numpy as np, numpy.random
>>> print np.random.dirichlet(np.ones(10),size=1)
[[ 0.01779975 0.14165316 0.01029262 0.168136 0.03061161 0.09046587
0.19987289 0.13398581 0.03119906 0.17598322]]
>>> print np.random.dirichlet(np.ones(10)/1000.,size=1)
[[ 2.63435230e-115 4.31961290e-209 1.41369771e-212 1.42417285e-188
0.00000000e+000 5.79841280e-143 0.00000000e+000 9.85329725e-005
9.99901467e-001 8.37460207e-246]]
>>> print np.random.dirichlet(np.ones(10)*1000.,size=1)
[[ 0.09967689 0.10151585 0.10077575 0.09875282 0.09935606 0.10093678
0.09517132 0.09891358 0.10206595 0.10283501]]
Depending on the main parameter the Dirichlet distribution will either give vectors where all the values are close to 1./N where N is the length of the vector, or give vectors where most of the values of the vectors will be ~0 , and there will be a single 1, or give something in between those possibilities.
EDIT (5 years after the original answer): Another useful fact about the Dirichlet distribution is that you naturally get it, if you generate a Gamma-distributed set of random variables and then divide them by their sum.
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