Your problem is that your 4/4 grid divides the quarters into four sixteenths, and, as you've noticed, sixteen is not divisible by three without a remainder. The subdivision of the quarter note must have a factor of three; the easiest, unless the tempo is very slow, is to subdivide them by three (into 8th note triplets). Equivalently, you can think of it as 12/8.

If the subdivision does not have three as one of its prime factors, then the triplet half notes will not coincide with the subdivision.

Then each of the triplet half notes in your example gets four of those subdivisions. When you do that, you just invert the first grid:

    _3_   _3_   _3_   _3_
   /   \ /   \ /   \ /   \
4 |x| | | |x| | | |x| | | |
4 |x| | |x| | |x| | |x| | |

Then you could count it

    _3_   _3_   _3_   _3_
   /   \ /   \ /   \ /   \
4 |x| | | |x| | | |x| | | |
4 |x| | |x| | |x| | |x| | |
   1     2 &   3   a 4
 

I chose the offbeat syllables based on the assumption that you count the triplet quarter notes as "1 and-a 2 and-a 3 and-a 4 and-a"; you can of course change them to suit.

So, in response to your edit, the answer is yes.

Your problem is that your 4/4 grid divides the quarters into four sixteenths, and, as you've noticed, sixteen is not divisible by three without a remainder. The subdivision of the quarter note must have three as one of its prime factors; if it does not, then the triplet half notes will not coincide with the subdivision.

The easiest factor to use, unless the tempo is very slow, is of course three (that is, subdivide the quarters into 8th note triplets). Equivalently, you can think of it as 12/8.

Then each of the triplet half notes in your example gets four of those subdivisions. Note that this preserves the normal relationship between half notes and eighth notes, that is, a factor of four, but within the parallel triplet universe.

When you do that, you just invert the first grid:

    _3_   _3_   _3_   _3_
   /   \ /   \ /   \ /   \
4 |x| | | |x| | | |x| | | |
4 |x| | |x| | |x| | |x| | |

Then you could count it

    _3_   _3_   _3_   _3_
   /   \ /   \ /   \ /   \
4 |x| | | |x| | | |x| | | |
4 |x| | |x| | |x| | |x| | |
   1     2 &   3   a 4
 

I chose the offbeat syllables based on the assumption that you count the triplet quarter notes as "1 and-a 2 and-a 3 and-a 4 and-a"; you can of course change them to suit:

     _3_   _3_   _3_   _3_
    /   \ /   \ /   \ /   \
 4 | | | | | | | | | | | | |
 4  1 & a 2 & a 3 & a 4 & a

So, in response to your edit, the answer is yes.