A sufficient condition to have a solution is when

There are 2 more missionaries than cannibals.

This is how:

One missionary enters the boat, then he ferries one missionary, one cannibal, one missionary, one cannibal, etc, finishing with one missionary.  I believe with only one more missionary than cannibals you cannot have arbitrarily large groups.

A sufficient condition to have a solution is when

There are 2 more missionaries than cannibals.

This is how:

One missionary enters the boat, then he ferries one missionary, one cannibal, one missionary, one cannibal, etc, finishing with one missionary. 

I believe with only one more missionary than cannibals you cannot have arbitrarily large groups.

But GoblinGuide did even better:

In fact with just one extra missionary there still is a solution.

The following diagram demonstrates his trick:

    MCMCMCM
            >CM>
    MCMCM            CM
            <C<
    MCMCMC            M
            >MC>
    MCMC            MCM
            <M<
    MCMCM            CM
            >CM>
    MCM            CMCM
            <C<
    MCMC            MCM
            >MC>
    MC            MCMCM
            <M<
    MCM            CMCM
            >CM>
    M            CMCMCM
             <C
    MC            MCMCM
            MC>
                MCMCMCM