Others have answered your question... use the 20 M
range. However, suppose your meter didn't have
any range bigger than 200k? You can still measure
your 1M resistor. First get a resistor in the
100k-200k range and measure it, then connect the
1M resistor to be measured in parallel. (Note, as
another post mentioned, not to touch the leads
during either reading.)
Now use the "resistors in parallel" equation in
reverse: Normally, Rpar = (R0 * Rx) / (R0 + Rx).
Rearranging, Rx = (R0 * Rpar) / (R0 - Rpar)
So if R0 = 100k and the parallel combo Rpar is
90k, the unknown is (100k * 90k) / (100k - 90k)
= 900k.
The main problem with this approach is that any uncertainty (i.e.
tolerance) in the known resistance creates a much greater degree of
uncertainty in the result.
E.g. if R0 = 100k +/- 5% (i.e. 95k-105k) and the parallel combo is 90k,
the unknown is between:
(95k * 90k) / (95k - 90k) = 1710k
(105k * 90k) / (105k - 90k) = 630k
This is so inaccurate as to be practically useless.
[With a 10% resistor, i.e. 90k-110k, it would be *completely* useless; the
90k lower bound means that the upper bound on the unknown would be infinity.]
With a 2% resistor, the figures are more reasonable, but still quite
inaccurate:
(98k * 90k) / (98k - 90k) = 1102k
(102k * 90k) / (102k - 90k) = 765k
=> 933k +/- 18%.
With a 1% resistor, the inaccuracy is tolerable for some purposes:
(99k * 90k) / (99k - 90k) = 990k
(101k * 90k) / (101k - 90k) = 826k
=> 908k +/- 9%.
Using a larger resistor (e.g. 180k, or 220k if that produces an in-range
result) would reduce the inaccuracy further.