After the primer is removed from the leading strand, how does DNA polymerase I add dNTPs without a 3'-OH?
The leading strand post-initiation
First consider the DNA replication fork after initiation has occurred.
The 3′-OH primer on the leading strand is the 3′-end of the strand of DNA being synthesized in the 5′ to 3′ direction (circled in the diagram below, modified from Berg, Biochemistry). It is not removed because there is no need to do so. (The primer on the lagging strand is only removed because it is RNA.) So the problem posed in the question does not arise.
Remember that the reason for Okazaki fragments is that there is no DNA 3'-OH primer for the lagging strand, so a temporary RNA 3′-OH primer is generated instead, RNA polymerase — unlike DNA polymerase — is able to copy DNA without a primer. There is no such problem for the leading strand.

The leading strand at the origin of replication
As @canadianer points out in a comment, the remarks above do not apply to the initiation of DNA replication, the single instance where the leading strand must have an RNA primer. In the case of prokaryotes such as Escherichia coli replication proceeds in both directions so that a short time after initiation one has a replication bubble as shown in the figure below (adapted from Russel, iGenetics) :

It can be seen that, at the origin, the 3′-OH of the DNA from the lagging strand of the other direction of replication (boxed in red) can act as a primer for the DNA synthesis function of DNA polymerase I, after its exonuclease activity has removed a nucleotide from the start of the RNA primer on the leading strand.
References
Berg et al. Ch.27
The Cell, Ch. 5
Sandwalk: Blog page
Footnote
There was a typo in the original question which caused confusion about which strand the poster was concerned with. This answer assumes the leading strand.